#include <bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define ll long long
#define ld long double
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 200005
#define mod 1000000007
using namespace std;
typedef pair<int, int> pii;
ll n, m, sum[Max << 2], add[Max << 2], f, a, b;
void pushup(ll rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void PushDown(int rt, int ln, int rn) {
    //ln,rn为左子树，右子树的数字数量。
    if(add[rt]) {
        //下推标记
        add[rt << 1] += add[rt];
        add[rt << 1 | 1] += add[rt];
        //修改子节点的Sum使之与对应的Add相对应
        sum[rt << 1] += add[rt] * ln;
        sum[rt << 1 | 1] += add[rt] * rn;
        //清除本节点标记
        add[rt] = 0;
    }
}
void build(ll l, ll r, ll rt) {
    if(l == r) {
        scanf("%lld", &sum[rt]);
        return;
    }
    ll m = (r + l) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    pushup(rt);
}
void Update(int L, int R, int C, int l, int r, int rt) { //L,R表示操作区间，l,r表示当前节点区间，rt表示当前节点编号
    if(L <= l && r <= R) { //如果本区间完全在操作区间[L,R]以内
        sum[rt] += C * (r - l + 1); //更新数字和，向上保持正确
        add[rt] += C; //增加Add标记，表示本区间的Sum正确，子区间的Sum仍需要根据Add的值来调整
        return ;
    }
    int m = (l + r) >> 1;
    PushDown(rt, m - l + 1, r - m); //下推标记
    //这里判断左右子树跟[L,R]有无交集，有交集才递归
    if(L <= m)
        Update(L, R, C, l, m, rt << 1);
    if(R >  m)
        Update(L, R, C, m + 1, r, rt << 1 | 1);
    pushup(rt);
}
void update(ll L, ll C, ll l, ll r, ll rt) {
    if(l == r) {
        sum[rt] = C;
        return ;
    }
    ll m = (l + r) >> 1;
    if(L <= m)
        update(L, C, l, m, rt << 1);
    else
        update(L, C, m + 1, r, rt << 1 | 1);
    pushup(rt);
}
ll query(ll L, ll R, ll l, ll r, ll rt) {
    if(L <= l && r <= R) {
        return sum[rt];
    }
    ll m = (r + l) >> 1;
    PushDown(rt, m - l + 1, r - m);
    ll ans = 0;
    if(L <= m)
        ans += query(L, R, l, m, rt << 1);
    if(R > m)
        ans += query(L, R, m + 1, r, rt << 1 | 1);
    return ans;
}
int main() {
    scanf("%lld%lld", &n, &m);
    build(1, n, 1);
    for(int i = 0; i < m; i++) {
        char c[10];
        scanf("%s%lld%lld", &c, &a, &b);
        if(c[0] == 'C') {
            cin >> f;
            Update(a, b, f, 1, n, 1);
        } else
            printf("%lld\n", query(a, b, 1, n, 1));
    }
    return 0;
}