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 JAVA):

Solution:

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int q = sc.nextInt();

		SqrtDecompositionRMQRUQ sq = new SqrtDecompositionRMQRUQ(n);
		sq.update(0, n, Integer.MAX_VALUE);
		for(int i = 0; i < q; i++){
			int type = sc.nextInt();

			if(type == 0){
				int s = sc.nextInt();
				int t = sc.nextInt();
				int x = sc.nextInt();

				sq.update(s, t + 1, x);
			}
			else {
				int s = sc.nextInt();
				int t = sc.nextInt();

				System.out.println(sq.findMin(s, t + 1));
			}

		}
	}

}

class SqrtDecompositionRMQRUQ {
	int N, K;
	long[] data;
	long[] bucketMin;
	long[] bucketLazyUpdate;
	static final int sqrtN = 512;
	static final int NIL = Integer.MIN_VALUE;

	public SqrtDecompositionRMQRUQ(int n) {
		N = n;
		K = (N + sqrtN - 1) / sqrtN;
		data = new long[K * sqrtN];
		bucketLazyUpdate = new long[K];
		Arrays.fill(bucketLazyUpdate, NIL);
		bucketMin = new long[K];
	}

	// [s, t)
	void update(int s, int t, int x) {
		for (int k = 0; k < K; ++k) {
			int l = k * sqrtN, r = (k + 1) * sqrtN;
			if (r <= s || t <= l)
				continue;
			if (s <= l && r <= t) {
				bucketLazyUpdate[k] = x;
				bucketMin[k] = x;
			} else {
				if(bucketLazyUpdate[k] != NIL){
					for(int i = l; i < r; i++){
						data[i] = bucketLazyUpdate[k];

					}
					bucketLazyUpdate[k] = NIL;
				}
				for (int i = Math.max(s, l); i < Math.min(t, r); ++i) {
					data[i] = x;
				}
				bucketMin[k] = Long.MAX_VALUE;
				for(int i = l; i < r; i++){
					bucketMin[k] = Math.min(bucketMin[k], data[i]);
				}
			}
		}
	}

	long findMin(int s, int t) {
		long result = Long.MAX_VALUE;
		for (int k = 0; k < K; ++k) {
			int l = k * sqrtN, r = (k + 1) * sqrtN;
			if (r <= s || t <= l)
				continue;
			if (s <= l && r <= t) {
				result = Math.min(result, bucketMin[k]);
			}
			else {
				if(bucketLazyUpdate[k] != NIL){
					result = Math.min(result, bucketLazyUpdate[k]);
				}
				else {
					for (int i = Math.max(s, l); i < Math.min(t, r); ++i) {
						result = Math.min(result, data[i]);
					}
				}
			}
		}

		return result;
	}

	long get(int s) {
		int k = s / sqrtN;
		if(bucketLazyUpdate[k] != NIL){
			return bucketLazyUpdate[k];
		}
		else {
			return data[s];
		}
	}
}