// Luogu.org P1438
// https://www.cnblogs.com/cytus/p/9590071.html
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
#define LL long long
const int N = 1e5 + 7;
int n, q, m;
LL a[N], tr[N << 2], add[N << 2];

//快读
int read() {
    int 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;
}

void modify(int x, int d) {
    for (x += m; x; x >>= 1) tr[x] += d; //单点修改，一路向上更新
}

void modify(int l, int r, int d) { //区间修改
    LL lc = 0, rc = 0, c = 1;
    for (l = l + m - 1, r = r + m + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
        tr[l] += d * lc, tr[r] += d * rc;

        if (~l & 1) add[l ^ 1] += d, tr[l ^ 1] += d * c, lc += c;
        if (r & 1) add[r ^ 1] += d, tr[r ^ 1] += d * c, rc += c;
    }
    //一路向上走,修改sum值，但却没有修改add
    for (; l; l >>= 1, r >>= 1) tr[l] += d * lc, tr[r] += d * rc;
}

LL query(int l, int r) {
    LL res = 0, lc = 0, rc = 0, c = 1;
    for (l = l + m - 1, r = r + m + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
        if (add[l]) res += add[l] * lc;
        if (add[r]) res += add[r] * rc;
        if (~l & 1) res += tr[l ^ 1], lc += c;
        if (r & 1) res += tr[r ^ 1], rc += c;
    }
    for (; l; l >>= 1, r >>= 1) res += add[l] * lc, res += add[r] * rc;
    return res;
}
int main() {
//文件输入输出
#ifndef ONLINE_JUDGE
    freopen("P1438.in", "r", stdin);
#endif
    n = read(), q = read();

    // build
    for (m = 1; m <= n + 1;) m <<= 1;
    for (int i = 1; i <= n; i++) {
        a[i] = read();
        tr[m + i] = a[i] - a[i - 1]; //记录差分值
    }

    // debug
    // printf("m=%d\n", m);
    // for (int i = 0; i <= 6; i++) printf("%lld ", tr[m + i]);
    // puts("");

    //所有父节点，从下向上，将值更新为左右子树的和,每个值都是自已所辖区间的差分和
    for (int i = m - 1; i >= 1; i--) {
        tr[i] = tr[i << 1] + tr[i << 1 | 1];
        // printf("tr[%d]=%d\n", i, tr[i]);
    }

    /*
    1 2 3 4 5

    1 2 3 4 5
      1 3 5
    ----------
    1 3 6 9 5

    query(3)=6

    tr
    1 1 1 1 1

    */

    for (int i = 1; i <= q; i++) {
        int op = read();
        if (op == 1) {
            int l = read(), r = read();
            LL k = read(), d = read(); //首项:k,公差:d
            // 1 2 4 1 2
            modify(l, k);                      //改2
            modify(r + 1, -(k + (r - l) * d)); //改4+1
            modify(l + 1, r, d);               //改3-4
        } else {
            LL x = read();                 //查询单点
            printf("%lld\n", query(1, x)); //以前缀和进行描述
        }
    }
    return 0;
}