Problem: Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:

* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.



Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Constraints

* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1

Input


n q
query1
query2
:
queryq


In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi 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 find operation, print the minimum value.

Examples

Input

3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2


Output

1
2


Input

1 3
1 0 0
0 0 0 5
1 0 0


Output

2147483647
5

Solution (in CPP):

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


bool upd[(1<<18)];
int val[(1<<18)];

inline void eval(int k,int r,int l){
	if( r-l>1 && upd[k] ){
		upd[k*2+1] = upd[k*2+2] = 1;  
		val[k*2+1] = val[k*2+2] = val[k];
	}
	upd[k]=0;
}

void update(int a,int b,int x,int k=0,int l=0,int r=(1<<17)){
	if(b<=l || r<=a)return;
	eval(k,r,l);

	if(a<=l && r<=b){
		upd[k] = 1;
		val[k] = x;
	}
	else{
		update(a,b,x,k*2+1,l,(l+r)/2);
		update(a,b,x,k*2+2,(l+r)/2,r);
		val[k] = min(val[k*2+1],val[k*2+2]);
	}
}
int find(int a,int b,int k=0,int l=0,int r=(1<<17)){
	if(b<=l || r<=a)return INT_MAX;
	eval(k,r,l);
	return (a<=l && r<=b)?val[k]:min( find(a,b,k*2+1,l,(l+r)/2) , find(a,b,k*2+2,(l+r)/2,r) );
}

int n,m,a,b,c,d;

int main(){
	fill(val,val+(1<<18),INT_MAX);
	cin>>n>>m;
	while(m--){
		cin>>a>>b>>c;
		if(a){
			cout<<find(b,c+1)<<endl;
		}
		else{
			cin>>d;
			update(b,c+1,d);
		}
	}
}