/*
题目：
将某区间每一个数加上k
求出某区间每一个数的和

题目总结：
区间修改，区间查询


*/

#include <iostream>
using namespace std;
#define LL long long
#define N 100010

//快读
LL read() {
    LL x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
int n, q, m;
LL tr[N << 2], tag[N << 2];

void build() {
    for (m = 1; m <= n;) m <<= 1;
    for (int i = 1; i <= n; i++) tr[m + i] = read();
    for (int i = m - 1; i >= 1; i--) tr[i] = tr[i << 1] + tr[i << 1 | 1];
}

void add(int l, int r, int d) {
    int lc = 0, rc = 0, c = 1;
    // 开区间，从左右叶子端点出发，如果l和r不是兄弟节点，就一路向上
    // lc 表示当前左端点走到的子树有多少个元素在修改区间内
    // rc 表示当前右端点走到的子树有多少个元素在修改区间内
    // c 当前层叶子节点个数
    for (l = m + l - 1, r = m + r + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
        tr[l] += d * lc, tr[r] += d * rc;                         //自底向上，更新统计信息tr和
        if (~l & 1) tag[l ^ 1] += d, tr[l ^ 1] += d * c, lc += c; // l % 2 == 0 左端点 是左儿子 右子树完整在区间内
        if (r & 1) tag[r ^ 1] += d, tr[r ^ 1] += d * c, rc += c;  // r % 2 == 1 右端点 是右儿子 左子树完整在区间内
    }
    //是兄弟节点时，需要一路向上到根进行推送
    for (; l || r; l >>= 1, r >>= 1) tr[l] += d * lc, tr[r] += d * rc;
}

//区间查询
LL query(int l, int r) {
    LL lc = 0, rc = 0, c = 1;
    LL res = 0;
    for (l = m + l - 1, r = m + r + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
        if (tag[l]) res += tag[l] * lc;
        if (tag[r]) res += tag[r] * rc;
        if (~l & 1) res += tr[l ^ 1], lc += c;
        if (r & 1) res += tr[r ^ 1], rc += c;
    }
    for (; l || r; l >>= 1, r >>= 1) res += tag[l] * lc, res += tag[r] * rc;
    return res;
}

int main() {
//文件输入输出
#ifndef ONLINE_JUDGE
    freopen("P3372.in", "r", stdin);
#endif
    n = read(), q = read();

    build();

    for (int i = 1; i <= q; i++) {
        int op = read();
        int l = read(), r = read();
        if (op == 1) {
            int c = read();
            add(l, r, c);
        } else
            printf("%lld\n", query(l, r));
    }
    return 0;
}