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<cstdio>
#include<algorithm>
#include<climits>
using namespace std;
static const int MAX_N = 100000;

int n, q;

int n_;
int dat[4 * MAX_N], laz[4 * MAX_N];

void init(){
	n_ = 1;
	while(n_ < n) n_ *= 2;
	for(int i = 0; i < 2 * n_ - 1; i++){
		dat[i] = INT_MAX;
		laz[i] = -1;
	}
}
void prop(int k){
	if(laz[k] == -1) return;
	if(2 * k + 1 < 2 * n_ - 1){
		laz[2 * k + 1] = laz[k];
		laz[2 * k + 2] = laz[k];
	}
	dat[k] = laz[k];
	laz[k] = -1;
}
void update(int a, int b, int x, int k, int l, int r){
	prop(k);
	if(r <= a || b <= l) return;
	if(a <= l && r <= b){
		laz[k] = x;
		return;
	}
	prop(k);
	update(a, b, x, 2 * k + 1, l, (l + r) / 2);
	update(a, b, x, 2 * k + 2, (l + r) / 2, r);
	int vl = (laz[2 * k + 1] == -1 ? dat[2 * k + 1] : laz[2 * k + 1]);
	int vr = (laz[2 * k + 2] == -1 ? dat[2 * k + 2] : laz[2 * k + 2]);
	dat[k] = min(vl, vr);
}
int find(int a, int b, int k, int l, int r){
	prop(k);
	if(r <= a || b <= l) return INT_MAX;
	if(a <= l && r <= b) return dat[k];
	int vl = find(a, b, 2 * k + 1, l, (l + r) / 2);
	int vr = find(a, b, 2 * k + 2, (l + r) / 2, r);
	return min(vl, vr);
}

int main(){
	scanf("%d %d", &n, &q);
	init();
	update(0, n + 1, INT_MAX, 0, 0, n_);
	for(int i = 0; i < q; i++){
		int op;
		scanf("%d", &op);
		if(op == 0){
			int s, t, x;
			scanf("%d %d %d", &s, &t, &x);
			update(s, t + 1, x, 0, 0, n_);
		}else{
			int s,t;
			scanf("%d %d", &s, &t);
			printf("%d\n", find(s, t + 1, 0, 0, n_));
		}
	}
	return 0;
}