Problem: Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:

* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.



Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.

Constraints

* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$

Input


$n$ $q$
$query_1$
$query_2$
:
$query_q$


In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:


0 $s$ $t$ $x$


or


1 $s$ $t$


The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.

Output

For each $getSum$ query, print the sum in a line.

Example

Input

6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5


Output

-5
1
6
8

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define FOR(i, a, b) for (int i = a; i <= (b); i++)
#define ROF(i, a, b) for (int i = a; i >= (b); i--)

using pii = pair<int, int>; using vpii = vector<pii>;
using vi = vector<int>; using vvi = vector<vi>;
using ll = long long;
using pll = pair<ll, ll>; using vpll = vector<pll>;
using vll = vector<ll>; using vvll = vector<vll>;

struct ruqst{
    vll t, d, tag;

    ruqst(int len, int init) {
        t.resize((len + 5) * 4, init);
        d.resize((len + 5) * 4, 0);
        tag.resize((len + 5) * 4, 0);
    }

    ll qry(int p, int l, int r, int tl, int tr) {
        if (tl <= l && r <= tr) return t[p];
        int mid = (l + r) / 2;
        if (tag[p]) {
            t[p * 2] = d[p] * (mid - l + 1), d[p * 2] = d[p], tag[p * 2] = 1;
            t[p * 2 + 1] = d[p] * (r - mid), d[p * 2 + 1] = d[p], tag[p * 2 + 1] = 1;
            d[p] = 0, tag[p] = 0;
        }
        ll rt = 0;
        if (tl <= mid) rt += qry(p * 2, l, mid, tl, tr);
        if (mid < r && tr > mid) rt += qry(p * 2 + 1, mid + 1, r, tl, tr);
        return rt;
    }

    void set(int p, int l, int r, int tl, int tr, ll x) {
        if (tl <= l && r <= tr) {
            t[p] = x * (r - l + 1), d[p] = x, tag[p] = 1;
            return;
        }
        int mid = (l + r) / 2;
        if (tag[p]) {
            t[p * 2] = d[p] * (mid - l + 1), d[p * 2] = d[p], tag[p * 2] = 1;
            t[p * 2 + 1] = d[p] * (r - mid), d[p * 2 + 1] = d[p], tag[p * 2 + 1] = 1;
            d[p] = 0, tag[p] = 0;
        }
        if (tl <= mid) set(p * 2, l, mid, tl, tr, x);
        if (mid < r && tr > mid) set(p * 2 + 1, mid + 1, r, tl, tr, x);
        t[p] = t[p * 2] + t[p * 2 + 1];
        return;
    }
};

int main() {
    // freopen("n", "r", stdin);
    // freopen("a.out", "w", stdout);
    int n, q; cin >> n >> q;
    ruqst st(n, 0);
    FOR(_, 1, q) {
        int c; cin >> c;
        if (c == 0) {
            int s, t, x; cin >> s >> t >> x;
            st.set(1, 0, n - 1, s, t, x);
        }
        else {
            int s, t; cin >> s >> t;
            cout << st.qry(1, 0, n - 1, s, t) << endl;
        }
    }
}