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 <algorithm>

constexpr int MAX_N = 100000, INF = 2147483647;

int n, q;

int seg[MAX_N * 4], lazy[MAX_N * 4];

bool flag[MAX_N * 4];

void init(int size) {

	n = 1;

	while (n < size)
		n  *= 2;

	for (int i = 0; i < n * 2 - 1; ++i) {
		seg[i] = INF;
		lazy[i] = INF;
	}
}

void eval(int k, int l, int r) {

	if (flag[k]) {

		seg[k] = lazy[k];

		if (r - l > 1) {
			lazy[k * 2 + 1] = lazy[k * 2 + 2] = lazy[k];
			flag[k * 2 + 1] = flag[k * 2 + 2] = true;
		}
	}
	flag[k] = false;
}

void update(int a, int b, int v, int k, int l, int r) {

	eval(k, l, r);

	if (r <= a || b <= l)
		return;

	if (a <= l&&r <= b) {

		lazy[k] = v;

		flag[k] = true;

		eval(k, l, r);

		return;
	}

	update(a, b, v, k * 2 + 1, l, (l + r) / 2);

	update(a, b, v, k * 2 + 2, (l + r) / 2, r);

	seg[k] = std::min(seg[k * 2 + 1], seg[k * 2 + 2]);
}

int query(int a, int b, int k, int l, int r) {
	
	eval(k, l, r);

	if (r <= a || b <= l)
		return INF;

	if (a <= l&&r <= b)
		return seg[k];

	return std::min(query(a, b, k * 2 + 1, l, (l + r) / 2), query(a, b, k * 2 + 2, (l + r) / 2, r));
}

int main() {

	std::cin >> n >> q;

	init(n);

	int com, s, t, x;

	for (int i = 0; i < q; ++i) {

		std::cin >> com >> s >> t;

		if (com == 0) {
			std::cin >> x;
			update(s, t + 1, x, 0, 0, n);
		}
		else
			std::cout << query(s, t + 1, 0, 0, n) << std::endl;
	}

	return 0;
}