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;

typedef long long ll;

struct lst
{
private:
    int n;
    vector<ll> node;
    vector<ll> lazy;
    vector<bool> lflg;

    void lazy2node(int k, int l, int r)
    {
        if(lflg[k])
        {
            node[k]=(r-l+1)*lazy[k];
            prop(lazy[k], k, l, r);
            lazy[k]=0;
            lflg[k]=false;
        }
    }

    void prop(ll v, int k, int l, int r)
    {
        if(r-l>0)
        {
            lazy[2*k+1]=lazy[2*k+2]=v;
            lflg[2*k+1]=lflg[2*k+2]=true;
        }
    }

public:
    lst(int sz)
    {
        n=1; while(n<sz) n*=2;
        node.resize(2*n-1, 0);
        lazy.resize(2*n-1, 0);
        lflg.resize(2*n-1, false);
    }
    void update(int qa, int qb, ll v, int k=0, int l=0, int r=-1)
    {
        if(r<0) r=n-1;
        lazy2node(k, l, r);
        if(qb<l || r<qa) return; 
        if(qa<=l && r<=qb)
        {
            node[k]=(r-l+1)*v;
            prop(v, k, l, r);
        }
        else
        {
            int m=(l+r)/2;
            update(qa, qb, v, 2*k+1, l, m);
            update(qa, qb, v, 2*k+2, m+1, r);
            node[k]=node[2*k+1]+node[2*k+2];
        }
    }

    ll get(int qa, int qb, int k=0, int l=0, int r=-1)
    {
        if(r<0) r=n-1;
        lazy2node(k, l, r);
        if(qb<l || r<qa) return 0;
        if(qa<=l && r<=qb) return node[k];
        else
        {
            int m=(l+r)/2;
            ll ls=get(qa, qb, 2*k+1, l, m);
            ll rs=get(qa, qb, 2*k+2, m+1, r);
            return ls+rs;
        }
    }
};

int main()
{
    int n, q;
    scanf("%d %d", &n, &q);
    lst tree=lst(n);
    for(int i=0; i<q; i++)
    {
        int c;
        scanf("%d", &c);
        if(c)
        {
            int s, t;
            scanf("%d %d", &s, &t);
            printf("%lld\n", tree.get(s,t));
        }
        else
        {
            int s, t, v;
            scanf("%d %d %d", &s, &t, &v);
            tree.update(s,t, (ll)v);
        }
    }
    return 0;
}