#include <bits/stdc++.h>
using namespace std;
int n, m;
struct node
{
    int data = 0, lazy = 0, ges = 0;
} dt[500005];
int num[100005];
int numpos = 0;
void build(int ll, int rr, int nowpos)
{
    dt[nowpos].ges = rr - ll + 1;
    if (ll == rr)
    {
        dt[nowpos].data = num[++numpos];
        return;
    }
    int mid = (ll + rr) >> 1;
    build(ll, mid, nowpos * 2);
    build(mid + 1, rr, nowpos * 2 + 1);
}
void add(int ql, int qr, int nowl, int nowr, int k, int nowpos)
{
    if (dt[nowpos].lazy != 0)
    { // 下放标记
        dt[nowpos * 2].data += dt[nowpos].lazy * dt[nowpos * 2].ges;
        dt[nowpos * 2].lazy += dt[nowpos].lazy;
        dt[nowpos * 2 + 1].data += dt[nowpos].lazy * dt[nowpos * 2 + 1].ges;
        dt[nowpos * 2 + 1].lazy += dt[nowpos].lazy;
        dt[nowpos].lazy = 0; // 清除当前节点的标记
    }
    int mid = (nowl + nowr) >> 1;
    int leftl = nowl, leftr = mid;
    int rghtl = mid + 1, rghtr = nowr;
    if (ql >= leftl && qr <= leftr)
    {
        if (ql == leftl && qr == leftr)
        {
            dt[nowpos * 2].lazy += k;
            dt[nowpos * 2].data += dt[nowpos * 2].ges * k;
            return;
        }
        else
            add(ql, qr, leftl, leftr, k, nowpos * 2);
    }
    else if (qr >= rghtl && qr <= rghtr)
    {
        if (ql == rghtl && qr == rghtr)
        {
            dt[nowpos * 2 + 1].lazy += k;
            dt[nowpos * 2 + 1].data += dt[nowpos * 2 + 1].ges * k;
            return;
        }
        else
            add(ql, qr, rghtl, rghtr, k, nowpos * 2 + 1);
    }
    else if (ql > leftl && ql < leftr && qr > rghtl && qr < rghtr)
    {
        add(ql, leftr, leftl, leftr, k, nowpos * 2);
        add(rghtl, qr, rghtl, rghtr, k, nowpos * 2 + 1);
    }
}
int query(int ql, int qr, int nowl, int nowr, int nowpos)
{
    if (dt[nowpos].lazy != 0)
    { // 下放标记
        dt[nowpos * 2].data += dt[nowpos].lazy * dt[nowpos * 2].ges;
        dt[nowpos * 2].lazy += dt[nowpos].lazy;
        dt[nowpos * 2 + 1].data += dt[nowpos].lazy * dt[nowpos * 2 + 1].ges;
        dt[nowpos * 2 + 1].lazy += dt[nowpos].lazy;
        dt[nowpos].lazy = 0; // 清除当前节点的标记
    }
    if (ql == nowl && qr == nowr)
        return dt[nowpos].data;
    int mid = (nowl + nowr) >> 1;
    int leftl = nowl, leftr = mid;
    int rghtl = mid + 1, rghtr = nowr;
    if (qr <= leftr)
        return query(ql, qr, leftl, leftr, nowpos * 2);
    else if (ql >= rghtl)
        return query(ql, qr, rghtl, rghtr, nowpos * 2 + 1);
    else
        return query(ql, leftr, leftl, leftr, nowpos * 2) + query(rghtl, qr, rghtl, rghtr, nowpos * 2 + 1);
}
signed main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> num[i];
    }
    build(1, n, 1);
    for (int i = 1; i <= m; i++)
    {
        int op, l, r, k;
        cin >> op >> l >> r;
        if (op == 1)
        {
            cin >> k;
            add(l, r, 1, n, k, 1);
        }
        else
        {
            cout << query(l, r, 1, n, 1) << endl;
        }
    }
}