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"
using namespace std;

long long INF = (1 << 31) - 1;

class LazySegmentTree {
private:
	long long Size;
	vector<long long> Node, Lazy;
	vector<bool> Flag;
public:
	LazySegmentTree(long long N) {
		Size = 1;
		while (Size < N) Size <<= 1;
		Node.assign(Size * 2, INF);
		Lazy.assign(Size * 2, INF);
		Flag.assign(Size * 2, false);
	}
	void Eval(long long K, long long L, long long R) {
		if (Flag[K]) {
			Node[K] = Lazy[K];
			if (R - L > 1) {
				Lazy[K * 2] = Lazy[K];
				Flag[K * 2] = true;
				Lazy[K * 2 + 1] = Lazy[K];
				Flag[K * 2 + 1] = true;
			}
			Lazy[K] = INF;
			Flag[K] = false;
		}
	}
	void Update(long long A, long long B, long long X, long long K, long long L, long long R) {
		Eval(K, L, R);
		if (B <= L || R <= A) return;
		if (A <= L && R <= B) {
			Lazy[K] = X;
			Flag[K] = true;
			Eval(K, L, R);
			return;
		}
		Update(A, B, X, K * 2, L, (L + R) / 2);
		Update(A, B, X, K * 2 + 1, (L + R) / 2, R);
		Node[K] = min(Node[K * 2], Node[K * 2 + 1]);
	}
	void Update(long long A, long long B, long long X) {
		Update(A, B, X, 1, 0, Size);
		return;
	}
	long long Query(long long A, long long B, long long K, long long L, long long R) {
		Eval(K, L, R);
		if (B <= L || R <= A) return INF;
		if (A <= L && R <= B) return Node[K];
		return min(Query(A, B, K * 2, L, (L + R) / 2), Query(A, B, K * 2 + 1, (L + R) / 2, R));
	}
	long long Query(long long A, long long B) {
		return Query(A, B, 1, 0, Size);
	}
};

int main() {
	long long N, Q;
	cin >> N >> Q;
	LazySegmentTree Seg(N);
	for (int i = 0; i < Q; i++) {
		long long Mode;
		cin >> Mode;
		if (Mode == 0) {
			long long A, B, X;
			cin >> A >> B >> X;
			Seg.Update(A, B + 1, X);
		}
		else {
			long long A, B;
			cin >> A >> B;
			cout << Seg.Query(A, B + 1) << endl;
		}
	}
}