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 <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;

#define NUM 2147483647

int N = 1;
int* data;

void init(int first_N){
	while(N < first_N)N *= 2;
}

void update(int update_left,int update_right,int new_value,int node_id,int node_left,int node_right){
	if(update_right < node_left || update_left > node_right)return;
	else if(update_left <= node_left && update_right >= node_right){
		data[node_id] = new_value;
	}else{
		if(data[node_id] >= 0){
			update(node_left,node_right,data[node_id],2*node_id+1,node_left,(node_left+node_right)/2);
			update(node_left,node_right,data[node_id],2*node_id+2,(node_left+node_right)/2+1,node_right);
			data[node_id] = -1;
		}
		update(update_left,update_right,new_value,2*node_id+1,node_left,(node_left+node_right)/2);
		update(update_left,update_right,new_value,2*node_id+2,(node_left+node_right)/2+1,node_right);
	}
}

int query(int search_left,int search_right,int node_id,int node_left,int node_right){
	if(search_right < node_left || search_left > node_right){
		return NUM;
	}else if(node_left <= search_left && node_right >= search_right && data[node_id] >= 0){
		return data[node_id];
	}else if(search_left <= node_left && search_right >= node_right && data[node_id] >= 0){
		return data[node_id];
	}else{
		if(data[node_id] != -1)return data[node_id];
		else{
			int left = query(search_left,search_right,2*node_id+1,node_left,(node_left+node_right)/2);
			int right = query(search_left,search_right,2*node_id+2,(node_left+node_right)/2+1,node_right);
			return min(left,right);
		}
	}
}

int main(){
	int first_N,Q;
	scanf("%d %d",&first_N,&Q);

	init(first_N);

	data = new int[270000];

	for(int i = 0; i <= 2*N-2; i++)data[i] = NUM;

	int command,left,right,value;

	for(int i = 0; i < Q; i++){
		scanf("%d",&command);

		if(command == 0){
			scanf("%d %d %d",&left,&right,&value);
			update(left,right,value,0,0,N-1);
		}else{
			scanf("%d %d",&left,&right);
			printf("%d\n",query(left,right,0,0,N-1));
		}
	}

}