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 rep(i,n) for(int i=0;i<(n);i++)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

#define MAX 500005

template<typename T>
struct SegTree {
  const T INF = numeric_limits<T>::max();
  int n;
  vector<T> dat, lazy;
  SegTree(int _n) : n(), dat(_n*4,0), lazy(_n*4,INF) {
    int x = 1;
    while (x < _n) x <<= 1;
    n = x;
  }
  void set(int i, int x) { dat[i+n-1] = x; }
  void build() { for (int k = n-2; k >= 0; k--) dat[k] = dat[k*2+1] + dat[k*2+2]; }

  void eval(int k, int p) {
    if (lazy[k] == INF) return;
    if (k < n-1) {
      lazy[k*2+1] = lazy[k];
      lazy[k*2+2] = lazy[k];
    }
    dat[k] = lazy[k] * p;
    lazy[k] = INF;
  }
  void update(int a, int b, int x, int k, int l, int r) {
    eval(k, r-l);
    if (r <= a || b <= l) return;
    if (a <= l && r <= b) {
      lazy[k] = x;
      eval(k, r-l);
    } else {
      int mid = (l+r)>>1;
      update(a,b,x,k*2+1,l,mid);
      update(a,b,x,k*2+2,mid,r);
      dat[k] = dat[k*2+1] + dat[k*2+2];
    }
  }
  void update(int a, int b, int x) { update(a,b,x,0,0,n); }

  T query(int a, int b, int k, int l, int r) {
    eval(k, r-l);
    if (r <= a || b <= l) return 0;
    if (a <= l && r <= b) return dat[k];
    int mid = (l+r)>>1;
    T vl = query(a,b,k*2+1,l,mid);
    T vr = query(a,b,k*2+2,mid,r);
    return vl + vr;
  }
  T query(int a, int b) { return query(a,b,0,0,n); }
  inline T operator[](int a) { return query(a,a+1); }
};

int main(int, char**)
{
  int n, q;
  cin >> n >> q;

  SegTree<ll> seg(n);

  vector<ll> ans;
  rep(i,q) {
    int command; cin >> command;
    if (command == 0) {
      int s, t, x;
      cin >> s >> t >> x;
      seg.update(s,t+1,x);
    } else if (command == 1) {
      int s, t; cin >> s >> t;
      ans.push_back(seg.query(s,t+1));
    }
  }
  for (auto a : ans) cout << a << endl;

  return 0;
}