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 <iostream>
#include <climits>
#include <vector>
#include <algorithm>
using namespace std;

void lazy_query(int k, vector<int>& node, vector<int>& lazy, int n) {
	if (lazy[k] == INT_MAX)return;
	node[k] = lazy[k];
	if (k < n - 1) {
		lazy[2 * k + 1] = lazy[k];
		lazy[2 * k + 2] = lazy[k];
	}
	lazy[k] = INT_MAX;
}

void update(int a, int b, int x, int k, int l, int r, vector<int>&node, vector<int>& lazy, int n) {
	lazy_query(k, node, lazy, n);
	if (r <= a || b <= l)return;
	if (a <= l && r <= b) {
		lazy[k] = x;
		lazy_query(k, node, lazy, n);
		return;
	}
	else {
		lazy_query(k, node, lazy, n);
		update(a, b, x, 2 * k + 1, l, (l + r) / 2, node, lazy, n);
		update(a, b, x, 2 * k + 2, (l + r) / 2, r, node, lazy, n);
		node[k] = min(node[2 * k + 1], node[2 * k + 2]);
		return;
	}
}

int find(int a, int b, int k, int l, int r, vector<int>&node, vector<int>& lazy, int n) {
	lazy_query(k, node, lazy, n);
	if (r <= a || b <= l)return INT_MAX;
	if (a <= l && r <= b) {
		return node[k];
	}
	else {
		lazy_query(k, node, lazy, n);
		int vl = find(a, b, 2 * k + 1, l, (l + r) / 2, node, lazy, n);
		int vr = find(a, b, 2 * k + 2, (l + r) / 2, r, node, lazy, n);
		node[k] = min(node[2 * k + 1], node[2 * k + 2]);
		return min(vr, vl);
	}
}

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	int n_, q;  cin >> n_ >> q;
	int n = 1;
	while (n < n_) n *= 2;
	vector<int> node(2 * n - 1, INT_MAX);
	vector<int> lazy(2 * n - 1, INT_MAX);

	for (int i = 0; i < q; i++) {
		int com; cin >> com;
		if (com == 0) {
			int s, t, x; cin >> s >> t >> x;
			update(s, t + 1, x, 0, 0, n, node, lazy, n);
		}
		else {
			int s, t; cin >> s >> t;
			cout << find(s, t + 1, 0, 0, n, node, lazy, n) << endl;
		}
	}
}