ДО A OK

#include <iostream>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <set>
#include <map>
#define pii pair <int, int>
#define pb(x) push_back(x)
using namespace std;
using ll = long long;
using ld = long double;
using db = double;
void cv(vector <int> &v) {
    for (auto x : v) cout << x << ' ';
    cout << "\n";
}

void cvl(vector <ll> &v) {
    for (auto x : v) cout << x << ' ';
    cout << "\n";
}


void cvv(vector <vector <int> > &v) {
    for (auto x : v) cv(x);
    cout << "\n";
}

void cvb(vector <bool> v) {
    for (bool x : v) cout << x << ' ';
    cout << "\n";
}

void cvs(vector <string>  v) {
    for (auto a : v) {
        cout << a << "\n";
    }
}

void cvp(vector <pii> a) {
    for (auto p : a) {
        cout << p.first << ' ' << p.second << "\n";
    }
    cout << "\n";
}

int n, N, L, R,k;


bool sh = 0;

struct tree{
    vector <ll> v;
    vector <ll> d;
    //d[id] - какое значение надо присвоить всем эелементам на отрезке в-ны id
    void bld(){
        cin >> n >> k;
        N = log2(n);
        if (pow(2, N) < n) N++;
        N = pow(2, N);
        v.assign(2 * N - 1, 0);
        d.assign(2 * N - 1, -1);
    }

    void seeT() {
        cout << "v\n";
        int cnt = -1;
        for (int i = 0; i <= log2(N); ++i) {
            for (int j = 0; j < pow(2, i); ++ j) {
                cnt ++;
                cout << v[cnt] << ' ';
            } cout << "\n";
        }
        cout << "\n";

        cout << "d\n";
        cnt = -1;
        for (int i = 0; i <= log2(N); ++i) {
            for (int j = 0; j < pow(2, i); ++ j) {
                cnt ++;
                cout << d[cnt] << ' ';
            } cout << "\n";
        }
        cout << "\n";
    }

    void push(int id, int l, int r) {
        if (sh) {
            cout << "push\n";
            cout << "l r = " << l << ' ' << r << "\n";
        }
        if (d[id] == -1) {
            return;
        }
        v[id] = d[id] * (r - l + 1);
        if (r > l) {
            d[2 * id + 1] = d[id];
            d[2 * id + 2] = d[id];
        }
        d[id] = -1;
    }

    ll get(int id, int l, int r) {
        push(id, l, r);
        if (l >= L && r <= R) {
            return v[id];
        }
        else if (max(l, L) <= min(r, R)) {
            int m = (l + r) / 2;
            return get(2 * id + 1, l, m) + get(2 * id + 2, m + 1, r);
        }
        return 0;
    }



    void upd(int id, int l, int r, ll x) {
        push(id, l, r);
        if (sh) {
            cout << "upd\n";
            cout << "l r = " << l << ' ' << r << "\n";
        }
        //не вошел
        if (r < L || l > R) {
            if (sh) {
                cout << "NO INTERSECTION\n";
            }
            return;
        }
        else if (l >= L && r <= R) {//полностью вошел
            if (sh) {
                cout << "COMPLETE intersetion\n";
            }
            d[id] = x;
            push(id, l, r);
            return;
        }
        //частичное пересечение
        if (sh) {
            cout << "part intersection\n";
        }

        int m = (l + r) / 2;
        upd(2 * id + 1, l, m, x);
        upd(2 * id + 2, m + 1, r, x);
        v[id] = v[id * 2 + 1] + v[id * 2 + 2];
    }
};
/*
5 9
A 2 3 2
A 3 5 1
A 4 5 2
Q 1 3
Q 2 2
Q 3 4
Q 4 5
Q 5 5
Q 1 5

*/

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    tree T;
    T.bld();
    for (int i = 0; i < k ; ++ i) {
        char q; cin >> q;
        cin >> L >> R;
        L--; R--;
        if (q == 'A') {
            ll x; cin >> x;
            T.upd(0, 0, N - 1, x);
        }
        else{
            cout << T.get(0, 0, N - 1) << "\n";
        }
        if (sh) {
            T.seeT();
        }
    }
}