#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10,M=1010;
int a[N],s1[N],s2[N],tag[N],n,B;
void pushdown(int x)
{
	int bl=(x-1)*B+1,br=min(x*B,n);
	for(int i=bl;i<=br;i++)a[i]=a[i]+tag[x];
	tag[x]=0;
}
void upd(int l,int r,int x)
{
	int bl=(l-1)/B+1,br=(r-1)/B+1;
	if(bl==br)
	{
		pushdown(bl);
		for(int i=l;i<=r;i++)
		{
			s1[bl]-=a[i]*a[i];
			a[i]+=x;s2[bl]+=x;
			s1[bl]+=a[i]*a[i];
		}
	}
	else
	{
		pushdown(bl);
		for(int i=l;i<=bl*B;i++)
		{
			s1[bl]-=a[i]*a[i];
			a[i]+=x;s2[bl]+=x;
			s1[bl]+=a[i]*a[i];
		}
		pushdown(br);
		for(int i=(br-1)*B+1;i<=r;i++)
		{
			s1[br]-=a[i]*a[i];
			a[i]+=x;s2[br]+=x;
			s1[br]+=a[i]*a[i];
		}
		for(int i=bl+1;i<br;i++)
		{
			s1[i]+=2*x*s2[i]+B*x*x;
			s2[i]+=B*x;
			tag[i]+=x;
		}
	}
}
int query(int l,int r)
{
	int bl=(l-1)/B+1,br=(r-1)/B+1,ans=0;
	if(bl==br)
	{
		pushdown(bl);
		for(int i=l;i<=r;i++)ans+=a[i]*a[i];
	}
	else
	{
		pushdown(bl);
		for(int i=l;i<=bl*B;i++)ans+=a[i]*a[i];
		pushdown(br);
		for(int i=(br-1)*B+1;i<=r;i++)ans+=a[i]*a[i];
		for(int i=bl+1;i<br;i++)ans+=s1[i];
	}
	return ans;
}
signed main()
{
	int q;cin>>n>>q;B=sqrt(n);
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		s1[(i-1)/B+1]+=a[i]*a[i];
		s2[(i-1)/B+1]+=a[i];
	}
	while(q--)
	{
		int op,l,r,c;cin>>op>>l>>r;
		if(op==0)cin>>c,upd(l,r,c);
		else cout<<query(l,r)<<'\n';
	}
	return 0;
}

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e6 + 10, sqrtN = 1050;
// 区间加，区间求平方和
// a[i]记录每个点的值，b[i]记录每个点i所在块;
// sum1[i]记录第i个块的一次方和，sum2[i]记录第i个块的平方和
// 每个块i的左端点L[i]、右端点R[i], 块内标记tag[i]
int n, m, a[N], b[N], L[sqrtN], R[sqrtN], tag[sqrtN], sum1[sqrtN], sum2[sqrtN];

void push_down(int x) { // 下传第x个块的标记
    if (tag[x]) {
        for (int i = L[x]; i <= R[x]; i++)
            a[i] += tag[x];
        tag[x] = 0;
    }
}

void calc(int x) { // 重新计算第x个块的sum1和sum2
    sum1[x] = 0;
    sum2[x] = 0;
    for (int i = L[x]; i <= R[x]; i++) {
        sum1[x] += a[i];
        sum2[x] += a[i] * a[i];
    }
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    int B = sqrt(n), cnt = (n + B - 1) / B; // B为每块的长度，cnt为总块数
    for (int i = 1; i <= n; i++) b[i] = (i - 1) / B + 1;
    for (int i = 1; i <= cnt; i++) {
        L[i] = (i - 1) * B + 1;
        R[i] = min(i * B, n);
        calc(i);
    }
    memset(tag, 0, sizeof(tag));
    for (int i = 1; i <= m; i++) {
        int op, l, r, c;
        cin >> op >> l >> r;
        if (op == 0) {
            cin >> c;
            if (b[l] == b[r]) { // 如果l和r在同一块内
                push_down(b[l]);
                for (int j = l; j <= r; j++) a[j] += c;
                calc(b[l]);
            } else {
                push_down(b[l]);
                for (int j = l; j <= R[b[l]]; j++) a[j] += c;
                calc(b[l]);

                for (int j = b[l] + 1; j <= b[r] - 1; j++) {
                    sum2[j] += 2 * c * sum1[j] + (R[j] - L[j] + 1) * c * c;
                    sum1[j] += (R[j] - L[j] + 1) * c;
                    tag[j] += c;
                }

                push_down(b[r]);
                for (int j = L[b[r]]; j <= r; j++) a[j] += c;
                calc(b[r]);
            }
        } else {
            int ans = 0;
            if (b[l] == b[r]) { // 如果l和r在同一块内
                for (int j = l; j <= r; j++) {
                    int val = a[j] + tag[b[l]];
                    ans += val * val;
                }
            } else {
                for (int j = l; j <= R[b[l]]; j++) {
                    int val = a[j] + tag[b[l]];
                    ans += val * val;
                }
                for (int j = b[l] + 1; j <= b[r] - 1; j++) {
                    ans += sum2[j];
                }
                for (int j = L[b[r]]; j <= r; j++) {
                    int val = a[j] + tag[b[r]];
                    ans += val * val;
                }
            }
            cout << ans << '\n';
        }
    }
    return 0;
}