#include<bits/stdc++.h>
using namespace std;
#define eb emplace_back
#define LL long long
template<typename T> void qr(T& x)
{
    char c=getchar(); x=0; int f=1;
    for(; !isdigit(c); c=getchar()) if(c=='-') f=-1;
    for(; isdigit(c); c=getchar()) x=x*10+(c-'0');
    x*=f;
}
template<typename T> void qw(T x)
{
    if(x<0) putchar('-'), x=-x;
    if(x/10) qw(x/10);
    putchar(x%10+'0'); 
}

const int N=1e6+10;
vector<int> G[N]; 
int n;LL v[N];

int fa[N], siz[N], dep[N], son[N];
void dfs1(int x, int xfa)
{
    fa[x]=xfa; 
    dep[x]=dep[xfa]+1; 

    siz[x]=1; 
    son[x]=0; 
    for(int y: G[x]) if(y!=xfa)
    {
        dfs1(y, x);
        siz[x]+=siz[y];
        if(siz[y]>siz[son[x]]) son[x]=y;
    }
}

int tsp, dfn[N], top[N];
void dfs2(int x, int tp) {
    dfn[x]=++tsp; top[x]=tp;
    if(son[x]) dfs2(son[x], tp);
    for(int y: G[x]) if(y!=son[x] && y!=fa[x])
        dfs2(y, y);
}

int LCA(int x,int y)
{
    for(;top[x]!=top[y];x=fa[top[x]])if(dep[top[x]]<dep[top[y]])swap(x, y);
	return dep[x]<dep[y] ? x : y;
}

LL c1[N], c2[N];
void add(LL c[], int x, LL k){if(x==0) return ;for(; x<=n; x+=x&-x) c[x]+=k;}
LL sum(LL c[], int x){LL res=0;for(; x>=1; x-=x&-x) res+=c[x];return res;}
LL getsum(LL c[], int l, int r){return sum(c, r)-sum(c, l-1);}

int main()
{
    int m,rt; qr(n),qr(m),qr(rt);
    for(int i=1; i<=n; i++)qr(v[i]);
    for(int i=1, x, y; i<n; i++) qr(x),qr(y),G[x].eb(y), G[y].eb(x);

    dep[0]=0;
    dfs1(rt, 0); 
    
    tsp=0;
    dfs2(rt, rt);

    for(int i=1; i<=n; i++) 
    {
        add(c1, dfn[i],      v[i]);   add(c2, dfn[i],      v[i]*dep[i]    );
        add(c1, dfn[fa[i]], -v[i]);   add(c2, dfn[fa[i]], -v[i]*dep[fa[i]]);
    }

    while(m--)
    {
        int op; scanf("%d", &op);
        if(op==1)      //1 x y k，表示将「结点 x 到结点 y 的简单路径」上所有结点的权值都增加 k
        {
            int x, y; LL k; qr(x); qr(y); qr(k);
            int lca=LCA(x, y), flca=fa[lca];
            add(c1,   dfn[x], k);    add(c2,   dfn[x],  k*dep[x]   );
            add(c1,   dfn[y], k);    add(c2,   dfn[y],  k*dep[y]   );
            add(c1, dfn[lca],-k);    add(c2, dfn[lca], -k*dep[lca] );
            add(c1,dfn[flca],-k);    add(c2,dfn[flca], -k*dep[flca]);
        }
        else if(op==2) //2 x，表示求结点 x 的权值
        {
            int x; qr(x);
            int l=dfn[x], r=dfn[x]+siz[x]-1;
            qw( getsum(c1, l, r) );
            printf("\n");
        }
        else           //3 x，表示求 x 的子树上所有结点的权值之和
        {
            int x; qr(x);
            int l=dfn[x], r=dfn[x]+siz[x]-1;
            qw( getsum(c2, l, r)-getsum(c1, l, r)*(dep[x]-1) );
            printf("\n");
        }
    }
    return 0;
}