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 mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define nyan "(=^･ω･^=)"
#define mkp make_pair
#define mkt make_tuple
#define lP pair<ll, ll>
#define iP pair<int,int>
typedef long long ll;
using namespace std;

int n, q, s, t, x, u, a[1 << 18], i;
bool f;
iP b[1 << 18];

void update(int k, int l, int r) {
	if (!(t < l || r < s)) {
		if (s <= l&&r <= t) {
			a[k] = x;
			b[k] = mkp(x, i);
		}
		else {
			if (b[k].second > b[k * 2].second) {
				b[k * 2] = b[k];
				a[k * 2] = b[k].first;
			}
			if (b[k].second > b[k * 2 + 1].second) {
				b[k * 2 + 1] = b[k];
				a[k * 2 + 1] = b[k].first;
			}
			update(k * 2, l, (l + r) / 2);
			update(k * 2 + 1, (l + r) / 2 + 1, r);
			a[k] = min(a[k * 2], a[k * 2 + 1]);
		}
	}
}
int find(int k, int l, int r) {
	if (t < l || r < s)return intmax;
	else if (s <= l&&r <= t)return a[k];
	else {
		if (b[k].second > b[k * 2].second) {
			b[k * 2] = b[k];
			a[k * 2] = b[k].first;
		}
		if (b[k].second > b[k * 2 + 1].second) {
			b[k * 2 + 1] = b[k];
			a[k * 2 + 1] = b[k].first;
		}
		return min(find(k * 2, l, (l + r) / 2), find(k * 2 + 1, (l + r) / 2 + 1, r));
	}
}
int main(){
	cin >> n >> q;
	for (i = 1; true; ++i) {
		if (1 << i >= n) {
			u = 1 << i;
			break;
		}
	}
	fill(a, a + u * 2, intmax);
	fill(b, b + u * 2, mkp(intmax, -1));
	for (i = 0; i != q; ++i) {
		cin >> f >> s >> t;
		if (f)
			cout << find(1, 0, u - 1) << endl;
		else {
			cin >> x;
			update(1, 0, u - 1);
		}
	}
	return 0;
}