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;
int INF = 2147483647;
template <typename T>
struct range_update_range_min{
	int N;
	vector<T> ST;
	vector<T> lazy;
	range_update_range_min(int n){
		N = 1;
		while (N < n){
			N *= 2;
		}
		ST = vector<T>(N * 2 - 1, INF);
		lazy = vector<T>(N * 2 - 1, -1);
	}
	range_update_range_min(vector<T> A){
		int n = A.size();
		N = 1;
		while (N < n){
			N *= 2;
		}
		ST = vector<T>(N * 2 - 1, INF);
		lazy = vector<T>(N * 2 - 1, -1);
		for (int i = 0; i < n; i++){
			ST[N - 1 + i] = A[i];
		}
		for (int i = N - 2; i >= 0; i--){
			ST[i] = min(ST[i * 2 + 1], ST[i * 2 + 2]);
		}
	}
	void eval(int i){
		if (lazy[i] != -1){
			if (i < N - 1){
				lazy[i * 2 + 1] = lazy[i];
				lazy[i * 2 + 2] = lazy[i];
			}
			ST[i] = lazy[i];
			lazy[i] = -1;
		}
	}
	void range_update(int L, int R, T x, int i, int l, int r){
		eval(i);
		if (R <= l || r <= L){
			return;
		} else if (L <= l && r <= R){
			lazy[i] = x;
			eval(i);
		} else {
			int m = (l + r) / 2;
			range_update(L, R, x, i * 2 + 1, l, m);
			range_update(L, R, x, i * 2 + 2, m, r);
			ST[i] = min(ST[i * 2 + 1], ST[i * 2 + 2]);
		}
	}
	void range_update(int L, int R, T x){
		return range_update(L, R, x, 0, 0, N);
	}
	T range_min(int L, int R, int i, int l, int r){
		eval(i);
		if (R <= l || r <= L){
			return INF;
		} else if (L <= l && r <= R){
			return ST[i];
		} else {
			int m = (l + r) / 2;
			return min(range_min(L, R, i * 2 + 1, l, m), range_min(L, R, i * 2 + 2, m, r));
		}
	}
	T range_min(int L, int R){
		return range_min(L, R, 0, 0, N);
	}
};
int main(){
  int n, q;
  cin >> n >> q;
  range_update_range_min<int> a(n);
  for (int i = 0; i < q; i++){
    int T;
    cin >> T;
    if (T == 0){
      int s, t, x;
      cin >> s >> t >> x;
      a.range_update(s, t + 1, x);
    }
    if (T == 1){
      int s, t;
      cin >> s >> t;
      cout << a.range_min(s, t + 1) << endl;
    }
  }
}