/*
a是原数组，b是a的差分数组, b[i] = a[i] - a[i - 1], a[i] = b[1] + b[2] + ... + b[i];

b1 b2 b3 ... bx
b1 b2 b3 ... bx
.
.
.
b1 b2 b3 ... bx
一共x+1项

求a的前缀和：
sum(a[1~x]) = b1 + (b1 + b2) + ... (b1 + b2 + ... + bx)
由上图的矩阵可知：
sum(a[1~x]) = sum(b[1~x] * (x+1) - (b1 + 2*b2 + 3*b3 + ... + x*bx)

因此可以用两个树状数组来分别维护 b 和 b[i]*i

*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;

const int N = 100010;

int n, m;
LL tr1[N];//维护差分数组
LL tr2[N];//维护b[i] * i的前缀和
int a[N];

int lowbit(int x)
{
    return x & -x;
}

void add(LL tr[], int x, LL c)
{
    for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}

LL sum(LL tr[], int x)
{
    LL res = 0;
    for (int i = x; i; i -= lowbit(i)) res += tr[i];
    return res;
}

LL prefix_sum(int x)
{
    return sum(tr1, x) * (x + 1) - sum(tr2, x);
}
int main()
{
    scanf("%d%d",&n, &m);
    for (int i = 1; i <= n; i ++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i ++)
    {
        int b = a[i] - a[i - 1];
        add(tr1, i, b);
        add(tr2, i, (LL)b * i);
    }

    while(m --)
    {
        char op[2];
        int l, r, d;
        scanf("%s%d%d",op, &l, &r);
        if (op[0] == 'Q')
        {
            printf("%lld\n", prefix_sum(r) - prefix_sum(l - 1));
        }
        else
        {
            scanf("%d", &d);
            // a[l] += d
            add(tr1, l, d), add(tr2, l, l * d);

            // a[r + 1] -= d; 
            //b[r+1]*(r+1) -> (b[r+1]-d)*(r+1) 即 b[r+1]*d - d*(r+1)
            add(tr1, r + 1, -d), add(tr2, r + 1, (r + 1) * -d);
        }
    }

    return 0;
}