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>
using namespace std;
const int inf = 2147483647;
int n, q, tp, v1, v2, v3, seg[262156], lazy[262156];
void push(int k) {
	if(lazy[k] != -inf) {
		seg[k * 2] = lazy[k * 2] = lazy[k];
		seg[k * 2 + 1] = lazy[k * 2 + 1] = lazy[k];
		lazy[k] = -inf;
	}
}
void update(int a, int b, int x, int k, int l, int r) {
	if (a <= l && r <= b) {
		seg[k] = lazy[k] = x;
		return;
	}
	if (r <= a || b <= l) return;
	push(k);
	update(a, b, x, k * 2, l, (l + r) >> 1);
	update(a, b, x, k * 2 + 1, (l + r) >> 1, r);
	seg[k] = min(seg[k * 2], seg[k * 2 + 1]);
}
int query(int a, int b, int k, int l, int r) {
	if (a <= l && r <= b) return seg[k];
	if (r <= a || b <= l) return inf;
	push(k);
	int lc = query(a, b, k * 2, l, (l + r) >> 1);
	int rc = query(a, b, k * 2 + 1, (l + r) >> 1, r);
	return min(lc, rc);
}
int main() {
	fill(seg, seg + 262156, inf);
	fill(lazy, lazy + 262156, inf);
	scanf("%d %d", &n, &q);
	while (q--) {
		scanf("%d", &tp);
		if (tp == 0) {
			scanf("%d %d %d", &v1, &v2, &v3); v2++;
			update(v1, v2, v3, 1, 0, 131072);
		}
		else {
			scanf("%d %d", &v1, &v2); v2++;
			int ret = query(v1, v2, 1, 0, 131072);
			printf("%d\n", ret);
		}
	}
	return 0;
}