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;
typedef long long ll;
const ll INF = ((ll)2<<30)-1;

ll dat[2<<18], add[2<<18];
int segn = 1;
void init(int n){
	while(segn < n)segn *= 2;
	for(int i = 0;i < segn*2;i++)dat[i] = INF;
	for(int i = 0;i < segn*2;i++)add[i] = -1;
}

void lazy_update(int k, int l, int r){
	if(add[k] == -1)return ;
	dat[k] = add[k];
	add[k*2+1] = add[k];
	add[k*2+2] = add[k];
	add[k] = -1;
}

void update(int a, int b, int x, int k, int l, int r){
	lazy_update(k, l, r);
	if(b <= l || r <= a)return ;
	if(a <= l && r <= b){
		add[k] = x;
		lazy_update(k, l, r);
		return ;
	}
	update(a, b, x, k*2+1, l, (l+r)/2);
	update(a, b, x, k*2+2, (l+r)/2, r);
	dat[k] = min(dat[k*2+1], dat[k*2+2]);
}

ll get(int a, int b, int k, int l, int r){
	lazy_update(k, l, r);
	if(b <= l || r <= a)return INF;
	if(a <= l && r <= b)return dat[k];
	ll vl = get(a, b, k*2+1, l, (l+r)/2);
	ll vr = get(a, b, k*2+2, (l+r)/2, r);
	return min(vl, vr);
}

int main(){
	int n, q, c, s, t, x;
	scanf("%d%d", &n, &q);
	init(n);
	for(int i = 0;i < q;i++){
		scanf("%d", &c);
		if(!c){
			scanf("%d%d%d", &s, &t, &x);
			update(s, t+1, x, 0, 0, segn);
		}else{
			scanf("%d%d", &s, &t);
			ll res = get(s, t+1, 0, 0, segn);
			printf("%lld\n", res);
		}
	}
	return 0;
}