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;
typedef long long LL;
struct SegTree {
	int size;
	LL *dat,*lazy;
	SegTree(int n) {
		size=1;
		while(size<n) size*=2;
		dat=(LL*)malloc(sizeof(LL)*(size*2-1));
		fill(dat,dat+(size*2-1),(1LL<<31)-1);
		lazy=(LL*)malloc(sizeof(LL)*(size*2-1));
		fill(lazy,lazy+(size*2-1),-1);
	}
	void eval(int k) {
		if(lazy[k]==-1) return;
		dat[k]=lazy[k];
		if(k<size-1) lazy[k*2+1]=lazy[k],lazy[k*2+2]=lazy[k];
		lazy[k]=-1;
	}
	void update(int a,int b,int  x,int k,int l,int r) {
		eval(k);
		if(r<=a||b<=l) return;
		if(a<=l&&r<=b) {
			lazy[k]=x;
			eval(k);
		}else {
			update(a,b,x,k*2+1,l,(l+r)/2);
			update(a,b,x,k*2+2,(l+r)/2,r);
			dat[k]=min(dat[k*2+1],dat[k*2+2]);
		}
	}
	LL getmin(int a,int b,int k,int l,int r) {
		eval(k);
		if(r<=a||b<=l) return (1LL<<31)-1;
		if(a<=l&&r<=b) return dat[k];
		return min(getmin(a,b,k*2+1,l,(l+r)/2),getmin(a,b,k*2+2,(l+r)/2,r));
	}
};
int main() {
	int n,q;
	scanf("%d %d",&n,&q);
	SegTree S(n);
	for(int i=0;i<q;i++) {
		int type,s,t,x;
		scanf("%d",&type);
		if(type==0) {
			scanf("%d %d %d",&s,&t,&x);t++;
			S.update(s,t,x,0,0,S.size);
		}else {
			scanf("%d %d",&s,&t);t++;
			printf("%lld\n",S.getmin(s,t,0,0,S.size));
		}
	}
}