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>
#include<algorithm>

using namespace std;

const long long INF = (1ll<<31)-1;

template<typename T>
class lazy_segment_tree{
	int n;
	T fval;
	vector<T> dat, lazy;
	vector<bool> lazyFlag;
	
public:
	
	lazy_segment_tree(){
		
	}
	
	lazy_segment_tree(int n_, T val){
		init(n_, val);
	}
	
	~lazy_segment_tree(){
		
	}
	
	void init(int n_, T val){
		fval = val;
		
		n = 1;
		
		while(n < n_) n *= 2;
		
		dat.resize(2 * n - 1);
		lazy.resize(2 * n - 1, fval);
		lazyFlag.resize(2 * n - 1, false);
	
		for(int i = 0; i < 2 * n - 1; i++) dat[i] = fval;
	}
	
	void eval(int k, int l, int r){
		
		if(lazyFlag[k]){
			dat[k] = lazy[k] * (r - l);
			
			if(r - l > 1) {
				lazy[2*k+1] = lazy[2*k+2] = lazy[k];
				lazyFlag[2*k+1] = lazyFlag[2*k+2] = lazyFlag[k];
			}
			
			lazyFlag[k] = false;
		}
	}
	
	void update(int a, int b, T x, int k = 0, int l = 0, int r = -1){
		if(r < 0) r = n;
		eval(k, l, r);
		if(b <= l || r <= a) {return ;}
		if(a <= l && r <= b){
			lazy[k] = x;
			lazyFlag[k] = true;
			eval(k, l, r);
		} else {
			update(a, b, x, 2*k+1, l, (l+r)/2);
			update(a, b, x, 2*k+2, (l+r)/2, r);
			dat[k] = (dat[2*k+1] + dat[2*k+2]);
		}
	}
	
	T query(int a, int b, int k = 0, int l = 0, int r = -1){
		if(r < 0) r = n;
		eval(k, l, r);
		if(r <= a || b <= l){
			return fval;
		}
		if(a <= l && r <= b){
			return dat[k];
		}else {
			T vl = query(a, b, k * 2 + 1, l, (l + r) / 2 );
			T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
			return (vl + vr);
		}
	}
};

signed main(){
	
	lazy_segment_tree<long long> lseg;
	
	int n, q;
	
	cin>>n>>q;
	
	lseg.init(n+10, 0);
	
	for(int i = 0; i < q; i++){
		int com, s, t, x;
		cin>>com>>s>>t;
		s++, t++;
		if(com){
			cout<<lseg.query(s,t+1)<<endl;
		} else {
			cin>>x;
			lseg.update(s, t+1, x);
		}
	}
	
	return 0;
}