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<iostream>
#include<vector>

using std::cin;
using std::cout;
using std::endl;

// Range Update, Range Sum
template<class T>
struct lazy_segment_tree{
  std::vector<T> data;
  std::vector<T> lazy;
  int n;
  T identity1,identity2;
  lazy_segment_tree(){}
  lazy_segment_tree(int sz,T ide1,T ide2):identity1(ide1),identity2(ide2){
    n = 1;
    while(n<sz)n<<=1;
    data.resize(2*n-1,identity1);
    lazy.resize(2*n-1,identity2);
  }
  lazy_segment_tree(std::vector<T> const& vec,T ide1,T ide2):identity1(ide1),identity2(ide2){
    int sz = vec.size();
    n = 1;
    while(n<sz)n<<=1;
    data.resize(2*n-1,identity1);
    lazy.resize(2*n-1,identity2);
    for(int i=0;i<sz;++i){
      data[i+n-1] = vec[i];
    }
    for(int i=sz-2;i>=0;--i){
      data[i] = data[i*2+1] + data[i*2+2];
    }
  }
  void eval(int k){
    if(lazy[k]==identity2)return;
    if(k<n-1){
      lazy[2*k+1] = lazy[k]/2;
      lazy[2*k+2] = lazy[k]/2;
    }
    data[k] = lazy[k];
    lazy[k] = identity2;
  }
  void update_impl(int a,int b,T x,int k,int l,int r){
    eval(k);
    if(a<=l && r<=b){
      lazy[k] = x*(r-l);
      eval(k);
    }
    else if(a<r && l<b){
      update_impl(a,b,x,2*k+1,l,(l+r)/2);
      update_impl(a,b,x,2*k+2,(l+r)/2,r);
      data[k] = data[2*k+1] + data[2*k+2];
    }
  }
  T query_impl(int a,int b,int k,int l,int r){
    eval(k);
    if(r<=a||b<=l){
      return identity1;
    }
    else if(a<=l && r<=b){
      return data[k];
    }
    else{
      T vl = query_impl(a,b,2*k+1,l,(l+r)/2);
      T vr = query_impl(a,b,2*k+2,(l+r)/2,r);
      return vl + vr;
    }
  }
  void update(int a,int b,T x){
    update_impl(a,b,x,0,0,n);
  }
  T query(int a,int b){
    T res = query_impl(a,b,0,0,n);
    return res;
  }
  T operator[](int i){
    return query(i,i+1);
  }
  void dump(){
    int cnt = 1;
    int now = 0;
    for(int i=0;i<2*n-1;++i){
      eval(i);
      cout<<data[i];
      now++;
      if(cnt==now){
        cout<<endl;
        cnt*=2;
        now=0;
      }else cout<<" ";
    }
  }
};

signed main(){
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  int N;cin>>N;
  lazy_segment_tree<int> data(N,0,1001);
  int Q;cin>>Q;
  while(Q--){
    int c;cin>>c;
    if(c==0){
      int s,t,x;cin>>s>>t>>x;
      data.update(s,t+1,x);
    }else{
      int s,t;cin>>s>>t;
      cout<< data.query(s,t+1) <<endl;
    }
    //data.dump();
  }
}