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++)
#define ALL(v) (v).begin(),(v).end()
typedef long long LL;
const LL INF = (1LL << 31) - 1;
const LL MOD = 1000000007LL;
using namespace std;
struct SegTree {
	LL *val;
	LL *lazy;
	int N;
	SegTree(int n) {
		N = 1;
		while (N < n) {
			N *= 2;
		}
		val = new LL[2 * N - 1];
		lazy = new LL[2 * N - 1];
		rep(i, 2 * N - 1) {
			val[i] = INF;
			lazy[i] = -1;
		}
	}
	void eval(int k) {
		if (lazy[k] != -1) {
			val[k] = lazy[k];
			if (k < N - 1) lazy[2 * k + 1] = lazy[k], lazy[2 * k + 2] = lazy[k];
		}
		lazy[k] = -1;
	}
	void update_q(int a, int b, LL x, int k, int l, int r) {
		eval(k);
		if (r <= a || b <= l) return;
		if (a <= l&&r <= b) {
			lazy[k] = x;
			eval(k);
			return;
		}
		if (k < N - 1) {
			update_q(a, b, x, k * 2 + 1, l, (l + r) / 2);
			update_q(a, b, x, k * 2 + 2, (l + r) / 2, r);
			val[k] = min(val[2 * k + 1], val[2 * k + 2]);
		}
	}
	LL min_q(int a, int b, int k, int l, int r) {
		eval(k);
		if (r <= a || b <= l) return INF;
		if (a <= l&&r <= b) return val[k];
		return min(min_q(a, b, k * 2 + 1, l, (l + r) / 2), min_q(a, b, k * 2 + 2, (l + r) / 2, r));
	}
};
int main() {
	int n, q;
	cin >> n >> q;
	SegTree S(n);
	rep(i, q) {
		int type;
		cin >> type;
		if (type) {
			int s, t;
			cin >> s >> t;
			cout << S.min_q(s, t + 1, 0, 0, S.N) << endl;
		}
		else {
			int s, t, x;
			cin >> s >> t >> x;
			S.update_q(s, t + 1, x, 0, 0, S.N);
		}
	}
}