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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef pair<int, int>P;

int dat[400000], lazy[400000], N;
inline void push(int k) {
	if (lazy[k] == -1)return;
	dat[k] = lazy[k];
	if (k < N - 1) {
		lazy[2 * k + 1] = lazy[k];
		lazy[2 * k + 2] = lazy[k];
	}
	lazy[k] = -1;
}
inline void update_node(int k) {
	dat[k] = min(dat[2 * k + 1], dat[2 * k + 2]);
}
inline void update(int a, int b, int x, int k, int l, int r) {
	push(k);
	if (r <= a || b <= l)return;
	if (a <= l&&r <= b) {
		lazy[k] = x; push(k); return;
	}
	update(a, b, x, k * 2 + 1, l, (l + r) / 2); update(a, b, x, k * 2 + 2, (l + r) / 2, r);
	update_node(k);
}
inline int query(int a, int b, int k, int l, int r) {
	push(k);
	if (r <= a || b <= l)return INT_MAX;
	if (a <= l&&r <= b)return dat[k];
	int lb = query(a, b, k * 2 + 1, l, (l + r) / 2);
	int rb = query(a, b, k * 2 + 2, (l + r) / 2, r);
	update_node(k);
	return min(lb, rb);
}
int main() {
	int n, q; scanf("%d%d", &n, &q);
	N = 1; while (N < n)N <<= 1;
	rep(i, 2 * N - 1)dat[i] = INT_MAX, lazy[i] = -1;
	rep(i, q) {
		int t; scanf("%d", &t);
		if (t == 0) {
			int a, b, x; scanf("%d%d%d", &a, &b, &x);
			update(a, b + 1, x, 0, 0, N);
		}
		else {
			int a, b; scanf("%d%d", &a, &b);
			printf("%d\n", query(a, b + 1, 0, 0, N));
		}
	}
}