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<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
#include<climits>

using namespace std;

const int INF=(1LL<<31)-1;
int N,K;
//const int sqrtN=512;
int sqrtN;
vector<int> data;
vector<int> bucketmin;
vector<int> lazyflag;
vector<int> lazyupdate;

void init(){
	K=(N+sqrtN-1)/sqrtN;
	data.assign(K*sqrtN,INF);
	bucketmin.assign(K,INF);
	lazyflag.assign(K,false);
	lazyupdate.assign(K,0);
}

void eval(int k){
	if(lazyflag[k]){
		lazyflag[k]=false;
		for(int i=k*sqrtN;i<(k+1)*sqrtN;i++){
			data[i]=lazyupdate[k];
		}
	}
}

void update(int s,int t, int x){
	for(int k=0;k<K;k++){
		int l=k*sqrtN;
		int r=(k+1)*sqrtN;
		if(r<=s || t<=l) continue;
		if(s<=l && r<=t){
			bucketmin[k]=x;
			lazyflag[k]=true;
			lazyupdate[k]=x;
		}
		else{
			eval(k);
			for(int i=max(s,l);i<min(t,r);i++){
				data[i]=x;
			}
			int &minval=bucketmin[k]=INF;
			for(int i=l;i<r;i++){
				minval=min(minval,data[i]);
			}
		}
	}
}

void find_min(int s,int t){
	int minval=INF;
	for(int k=0;k<K;k++){
		int l=k*sqrtN;
		int r=(k+1)*sqrtN;
		if(r<=s || t<=l) continue;
		if(s<=l && r<=t){
			minval=min(minval,bucketmin[k]);
		}
		else{
			eval(k);
			for(int i=max(s,l);i<min(t,r);i++){
				minval=min(minval,data[i]);
			}
		}
	}
	printf("%d\n",minval);
}

int main(){
	int q,com;
	int s,t,x;
	scanf("%d %d",&N,&q);
	sqrtN = (int)sqrt(N);
	//printf("sqrt(N)=%lld\n",sqrtN);
	init();
	for(int i=0;i<q;i++){
		scanf("%d",&com);
		if(com==0){
			scanf("%d %d %d",&s,&t,&x);
			update(s,t+1,x);//[s,t)
		}
		else{
			scanf("%d %d",&s,&t);
			find_min(s,t+1);
		}
	}
}