//对于操作1，明显就是维护一下差分数组bi
//对于操作2，推公式可以得出：=∑（1~x）∑（1~i）bj = ∑（1~x）bi * （x + 1）- ∑（1~x）i * bi
//所以只要维护两个前缀和：bi 和 ibi 就好了

#include<bits/stdc++.h>

#define lowbit(x) x & -x

using namespace std;
typedef long long ll;
const int N = 2e5 + 10, inf = 0x3f3f3f3f;

int n, m;
ll a[N], d[N];
ll tr1[N], tr2[N];//tr1是bi前缀和，tr2是i*bi的前缀和

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

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

ll qurry(ll x)
{
    return sum(tr1, x) * (x + 1) - sum(tr2, x);
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);

    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    for(int i = n; i; i --) d[i] = a[i] - a[i - 1], add(i, d[i], tr1), add(i, i * d[i], tr2);

    while(m --)
    {
        char c;
        cin >> c;

        if(c == 'C')
        {
            ll l, r, x;
            cin >> l >> r >> x;
            add(l, x, tr1), add(r + 1, -x, tr1);
            add(l, l * x, tr2), add(r + 1, -x * (r + 1), tr2);
        }
        else 
        {
            ll l, r;
            cin >> l >> r;
            cout << qurry(r) - qurry(l - 1) << '\n';
        }
    }
    return 0;
}