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;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int MOD = (int)(1e9) + 7;
const int INF = (1ll << 31) - 1;
const double PI = acos(-1);
const double EPS = 1e-9;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a > b) { a = b; return true; } return false; }
const int NIL = -1;
struct Seg {
	int n;
	vector<int>d, lazy;
	Seg(int m) {
		for (n = 1; n < m; n <<= 1);
		d.assign(2 * n, INF);
		lazy.assign(2 * n, NIL);
	}
	void eval(int k, int l, int r) {
		if (lazy[k] != NIL) {
			d[k] = lazy[k];
			if (r - l > 1) {
				lazy[2 * k] = lazy[2 * k + 1] = lazy[k];
			}
			lazy[k] = NIL;
		}
	}
	void update(int a, int b, int x, int k = 1, int l = 0, int r = -1) {
		if (r == -1)r = n;
		eval(k, l, r);
		if (r <= a || b <= l)return;
		else if (a <= l&&r <= b) {
			lazy[k] = x;
			eval(k, l, r);
		}
		else {
			update(a, b, x, 2 * k, l, (l + r) / 2);
			update(a, b, x, 2 * k + 1, (l + r) / 2, r);
			d[k] = min(d[2 * k], d[2 * k + 1]);
		}
	}
	int find(int a, int b, int k = 1, int l = 0, int r = -1) {
		if (r == -1)r = n;
		eval(k, l, r);
		if (r <= a || b <= l)return INF;
		else if (a <= l&&r <= b)return d[k];
		else return min(find(a, b, 2 * k, l, (l + r) / 2), find(a, b, 2 * k + 1, (l + r) / 2, r));
	}
};
signed main() {
	cin.tie(0);
	int n, q; cin >> n >> q;
	Seg seg(n);
	rep(i, 0, q) {
		int com; cin >> com;
		if (com) {
			int s, t; cin >> s >> t;
			cout << seg.find(s, t+1) << endl;
		}
		else {
			int s, t, x; cin >> s >> t >> x;
			seg.update(s, t+1, x);
		}
	}
	return 0;
}