先找一个判定条件。

设 $s_u$ 表示 $u$ 的子树中总共进行了多少次操作。对于节点 $u$，记 $S=\sum_{v\in son(u)}s_v$，即所有儿子子树一共进行了 $S$ 次操作。

定义 $r_u=(d_u-c_u)\bmod d_u$。

当 $S$ 确定后，节点 $u$ 自己可以额外操作 $x_u$ 次，其中 $0\le x_u\le r_u+(d_u-1)S$。

因此 $s_u=S+x_u$，所以有 $s_u\in[S,d_uS+r_u]$。

于是可以做树形 DP。设 $f_u[x]$ 表示 $s_u=x$ 时的方案数。对于节点 $u$，先把所有儿子的 DP 数组卷积起来，得到所有儿子的操作次数之和为 $S$ 的方案数，然后将 $S$ 转移到整个区间 $[S,d_uS+r_u]$。

考虑生成函数。定义 $F_u(x)=\sum_i f_u[i]x^i$，并设所有儿子的 GF 乘积为 $F(x)=\prod_{v\in son(u)}F_v(x)$。

若儿子的操作次数之和为 $S$，那么它对 $F_u$ 的贡献为 $x^S+x^{S+1}+\cdots+x^{d_uS+r_u}$，即 $\frac{x^S-x^{d_uS+r_u+1}}{1-x}$。

因此可以得到转移：

最终答案为 $F_1(1)$。但这个式子在 $x=1$ 处不方便计算，因此做代换 $x\to e^x$。

定义 $G_u(x)=F_u(e^x)$，同时令 $G(x)=F(e^x)$，则有 $F(e^{d_ux})=G(d_ux)$，于是转移变为

原来的答案 $F_1(1)$ 也变成了 $G_1(0)$，所以最终只需要求根节点生成函数的常数项。

又因为 $1-e^x=-x+O(x^2)$，每进行一次除以 $1-e^x$ 的操作，只会使我们需要的项数增加 $1$。因此，对于 $u$，就只需要保留 $G_u(x)$ 的 dep 项。

使用 NTT 优化，时间复杂度 $O(n^2\log n)$。

#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
#define ull unsigned long long
#define SZ(x) ((int)((x).size()))
#define ALL(x) (x).begin(),(x).end()
using namespace std;

inline int read(){
	char c=getchar();int x=0;bool f=0;
	for(;!isdigit(c);c=getchar())f^=!(c^45);
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
	return f?-x:x;
}

#define mod 998244353
struct modint{
	unsigned int x;
	modint(int o=0){x=o;}
	modint &operator = (int o){return x=o,*this;}
	modint &operator +=(modint o){return x=x+o.x>=mod?x+o.x-mod:x+o.x,*this;}
	modint &operator -=(modint o){return x=x<o.x?x-o.x+mod:x-o.x,*this;}
	modint &operator *=(modint o){return x=1ull*x*o.x%mod,*this;}
	modint &operator ^=(int b){
		modint a=*this,c=1;
		for(;b;b>>=1,a*=a)if(b&1)c*=a;
		return x=c.x,*this;
	}
	modint &operator /=(modint o){return *this *=o^=mod-2;}
	friend modint operator +(modint a,modint b){return a+=b;}
	friend modint operator -(modint a,modint b){return a-=b;}
	friend modint operator *(modint a,modint b){return a*=b;}
	friend modint operator /(modint a,modint b){return a/=b;}
	friend modint operator ^(modint a,int b){return a^=b;}
	friend bool operator ==(modint a,modint b){return a.x==b.x;}
	friend bool operator !=(modint a,modint b){return a.x!=b.x;}
	bool operator ! () {return !x;}
	modint operator - () {return x?mod-x:0;}
	bool operator <(const modint&b)const{return x<b.x;}
};
inline modint qpow(modint x,int y){return x^y;}

vector<modint> fac,ifac,iv;
inline void initC(int n)
{
	if(iv.empty())fac=ifac=iv=vector<modint>(2,1);
	int m=iv.size(); ++n;
	if(m>=n)return;
	iv.resize(n),fac.resize(n),ifac.resize(n);
	For(i,m,n-1){
		iv[i]=iv[mod%i]*(mod-mod/i);
		fac[i]=fac[i-1]*i,ifac[i]=ifac[i-1]*iv[i];
	}
}
inline modint C(int n,int m){
	if(m<0||n<m)return 0;
	return initC(n),fac[n]*ifac[m]*ifac[n-m];
}
inline modint sign(int n){return (n&1)?(mod-1):(1);}

#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int,int>pii;
typedef vector<int>vi;

#define poly vector<modint>
const modint G=3,Ginv=modint(1)/3;
inline poly one(){poly a;a.push_back(1);return a;}
vector<int>rev;
int rts[2100000];
inline int ext(int n){
	int k=0;
	while((1<<k)<n)++k;return k; 
}
inline void init(int k){
	int n=1<<k;
	rts[0]=1,rts[1<<k]=qpow(31,1<<(21-k)).x;
	Rep(i,k,1)rts[1<<(i-1)]=1ull*rts[1<<i]*rts[1<<i]%mod;
	For(i,1,n-1)rts[i]=1ull*rts[i&(i-1)]*rts[i&-i]%mod;
}

void ntt(poly&a,int k,int typ){
	int n=1<<k;
	static ull tmp[2100000];
	for(int i=0;i<n;++i)tmp[i]=a[i].x;
	if(typ==1){
		for(int l=n>>1;l>=1;l>>=1){
			ull*k=tmp;
			for(int*g=rts;k<tmp+n;k+=(l<<1),++g){
				for(ull*x=k;x<k+l;++x){
					int o=x[l]%mod*(*g)%mod;
					x[l]=*x+mod-o,*x+=o;
				}
			}
		}
		for(int i=0;i<n;++i)a[i].x=tmp[i]%mod;
	}else{
		for(int l=1;l<n;l<<=1){
			ull*k=tmp;
			for(int*g=rts;k<tmp+n;k+=(l<<1),++g){
				for(ull*x=k;x<k+l;++x){
					int o=x[l]%mod;
					x[l]=(*x+mod-o)*(*g)%mod,*x+=o;
				}
			}
		}
		int iv=qpow(n,mod-2).x;
		for(int i=0;i<n;++i)a[i].x=tmp[i]%mod*iv%mod;
		reverse(a.begin()+1,a.end());
	}
}
 
poly operator +(poly a,poly b){
	int n=max(a.size(),b.size());a.resize(n),b.resize(n);
	For(i,0,n-1)a[i]+=b[i];return a;
}
poly operator -(poly a,poly b){
	int n=max(a.size(),b.size());a.resize(n),b.resize(n);
	For(i,0,n-1)a[i]-=b[i];return a;
}
poly operator *(poly a,modint b){
	int n=a.size();
	For(i,0,n-1)a[i]*=b;return a;
} 
poly operator *(poly a,poly b){
	if(!a.size()||!b.size())return {};
	if((int)a.size()<=32 || (int)b.size()<=32){
		poly c(a.size()+b.size()-1,0);
		for(int i=0;i<a.size();++i)for(int j=0;j<b.size();++j)c[i+j]+=a[i]*b[j];
		return c; 
	}
	int n=(int)a.size()+(int)b.size()-1,k=ext(n);
	a.resize(1<<k),b.resize(1<<k);
	ntt(a,k,1),ntt(b,k,1);
	For(i,0,(1<<k)-1)a[i]*=b[i];
	ntt(a,k,-1),a.resize(n);return a;
}

poly Tmp;
poly pmul(poly a,poly b,int n,bool ok=0)
{
	int k=ext(n);
	a.resize(1<<k),ntt(a,k,1);
	if(!ok) b.resize(1<<k),ntt(b,k,1),Tmp=b;
	For(i,0,(1<<k)-1)a[i]*=Tmp[i];
	ntt(a,k,-1),a.resize(n);
	return a;
}
poly inv(poly a,int n)
{
	a.resize(n);
	if(n==1){
		poly f(1,1/a[0]);
		return f;
	}
	poly f0=inv(a,(n+1)>>1),f=f0;
	poly now=pmul(a,f0,n,0);
	for(int i=0;i<f0.size();++i)now[i]=0;
	now=pmul(now,poly(0),n,1);
	f.resize(n);
	for(int i=f0.size();i<n;++i)f[i]=-now[i];
	return f;
}
poly inv(poly a){return inv(a,a.size());}

poly deriv(poly a){
	int n=(int)a.size()-1;
	For(i,0,n-1)a[i]=a[i+1]*(i+1);
	a.resize(n);return a;
}
poly inter(poly a){
	int n=a.size()+1;a.resize(n); initC(n);
	Rep(i,n-1,1)a[i]=a[i-1]*iv[i];
	a[0]=0;return a;
}
poly ln(poly a){
	int n=a.size();
	a=deriv(a)*inv(a),a.resize(n-1);return inter(a);
}
poly exp(poly a,int k){
	int n=1<<k;a.resize(n);
	if(n==1)return one();
	poly f0=exp(a,k-1);f0.resize(n);
	return f0*(one()+a-ln(f0)); 
}
poly exp(poly a){
	int n=a.size();
	a=exp(a,ext(n));a.resize(n);return a;
}
poly div(poly a,poly b){
	int n=a.size(),m=b.size(),k=ext(n-m+1);
	reverse(a.begin(),a.end()),reverse(b.begin(),b.end());
	a.resize(n-m+1),b.resize(n-m+1);
	a=a*inv(b),a.resize(n-m+1),reverse(a.begin(),a.end()); return a;
}
poly modulo(poly a,poly b){
	if(b.size()>a.size())return a;
	int n=b.size()-1;
	a=a-div(a,b)*b;a.resize(n);return a;
}

#define maxn 200005
#define inf 0x3f3f3f3f

int n,c[maxn],d[maxn],fa[maxn],dep[maxn];
vi e[maxn],ord;
poly f[maxn],ber;

inline poly mul_lim(poly a,poly b,int lim)
{
	if(a.empty()||b.empty()||lim<=0)return {};
	if(SZ(a)>lim)a.resize(lim);
	if(SZ(b)>lim)b.resize(lim);
	if(SZ(a)<=32||SZ(b)<=32){
		poly res(min(lim,SZ(a)+SZ(b)-1),0);
		For(i,0,SZ(a)-1){
			int R=min(SZ(b)-1,lim-1-i);
			For(j,0,R)res[i+j]+=a[i]*b[j];
		}
		return res;
	}
	poly res=a*b;
	if(SZ(res)>lim)res.resize(lim);
	return res;
}

signed main()
{
	n=read();
	For(i,1,n)c[i]=read(),d[i]=read();
	For(i,1,n-1){
		int u=read(),v=read();
		e[u].pb(v),e[v].pb(u);
	}
	
	ord.pb(1),fa[1]=0,dep[1]=0;
	for(int p=0;p<SZ(ord);++p){
		int u=ord[p];
		for(int v:e[u])if(v!=fa[u]){
			fa[v]=u;
			dep[v]=dep[u]+1;
			ord.pb(v);
		}
	}
	
	int D=0;
	For(i,1,n)D=max(D,dep[i]);
	initC(D+2);
	init(ext(2*D+5));
	
	ber.resize(D+1);
	ber[0]=1;
	For(k,1,D){
		modint s=0;
		For(i,1,k)s+=ifac[i+1]*ber[k-i];
		ber[k]=-s;
	}
	
	Rep(p,SZ(ord)-1,0){
		int u=ord[p];
		int r=dep[u],lim=r+2;
		
		poly h(1,1);
		for(int v:e[u])if(fa[v]==u){
			h=mul_lim(h,f[v],lim);
		}
		h.resize(lim);
		
		poly hd(lim);
		modint pw=1;
		For(i,0,lim-1){
			hd[i]=h[i]*pw;
			pw*=d[u];
		}
		
		ll a=(d[u]-1ll*(c[u]%d[u]))%d[u];
		modint A=a+1;
		poly ex(lim);
		ex[0]=1;
		For(i,1,lim-1)ex[i]=ex[i-1]*A*iv[i];
		
		poly q=mul_lim(ex,hd,lim);
		q.resize(lim);
		For(i,0,lim-1)q[i]-=h[i];
		
		poly g(r+1);
		For(i,0,r)g[i]=q[i+1];
		
		poly b(r+1);
		For(i,0,r)b[i]=ber[i];
		
		f[u]=mul_lim(g,b,r+1);
		f[u].resize(r+1);
	}
	
	printf("%u\n",f[1][0].x);
	return 0;
}