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;
#define rep(i,n) for(int i=0;i<n;i++)
#define rrep(i,n) for(int i=1;i<=n;i++)
#define drep(i,n) for(int i=n;i>=0;i--)
#define MAX 131072
#define B 100
#define dpb push_back
#define fi first
#define se second
typedef pair<int, int> P;
typedef long long ll;
const int inf = 1e9;
const ll INF = 2147483647;
//__gcd(a,b), __builtin_popcount(a);

int n;
ll dat[MAX*2], add[MAX*2];
void init(int m){
	n = 1;
	while(n < m)n *= 2;
	for(int i = 0;i < 2*n;i++)dat[i] = INF;
	for(int i = 0;i < 2*n;i++)add[i] = -1;
}

void lazy_update(int k){
	if(add[k] == -1)return;
	if(k < n-1){
		add[k*2+1] = add[k];
		add[k*2+2] = add[k];
	}
	add[k] = -1;
}

void update(int a, int b, int x, int k, int l, int r){
	if(r <= a || b <= l)return;
	if(a <= l && r <= b){
		add[k] = dat[k] = x;
		lazy_update(k);
		return;
	}
	if(add[k] != -1)dat[k] = add[k];
	lazy_update(k);
	update(a, b, x, k*2+1, l, (l+r)/2);
	update(a, b, x, k*2+2, (l+r)/2, r);
	ll t1 = dat[k*2+1], t2 = dat[k*2+2];
	if(add[k*2+1] != -1)t1 = add[k*2+1];
	if(add[k*2+2] != -1)t2 = add[k*2+2];
	dat[k] = min(t1, t2);
	return;
}

ll find(int a, int b, int k, int l, int r){
	if(r <= a || b <= l)return INF;
	if(a <= l && r <= b){
		if(add[k] != -1){
			dat[k] = add[k];
		}
		lazy_update(k);
		return dat[k];
	}
	if(add[k] != -1)dat[k] = add[k];
	lazy_update(k);
	ll vl = find(a, b, k*2+1, l, (l+r)/2);
	ll vr = find(a, b, k*2+2, (l+r)/2, r);
	ll t1 = dat[k*2+1], t2 = dat[k*2+2];
	if(add[k*2+1] != -1)t1 = add[k*2+1];
	if(add[k*2+2] != -1)t2 = add[k*2+2];
	return min(vr, vl);
}

int main(){
	int m, q, c, s, t, x;
	scanf("%d%d", &m, &q);
	init(m);
	while(q--){
		scanf("%d", &c);
		if(c == 0){
			scanf("%d%d%d", &s, &t, &x);
			update(s, t+1, x, 0, 0, n);
		}else{
			scanf("%d%d", &s, &t);
			ll ans = find(s, t+1, 0, 0, n);
			printf("%lld\n", ans);
		}
	}
	/*rep(i,2*n-1)printf("%d ", dat[i]);
	printf("\n");
	rep(i,2*n-1)printf("%d ", add[i]);*/
	return 0;
}