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<(int)N;++i)
#define rep1(i,N) for(int i=1;i<=(int)N;++i)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define SZ(x) (int)x.size()
#define e_b emplace_back
using ll = long long;
const ll LINF=1LL<<60;
const int INF=1<<30;

struct LazySegmentTree{
	const ll INF=(1LL<<31)-1;
	int n;
	vector<ll> node,lazy;
	
	LazySegmentTree(int sz){
		n=1;
		while(sz>n) n<<=1;
		node.assign(2*n,INF);
		lazy.assign(2*n,-1);
	}
	
	void eval(int k,int l,int r){
		if(lazy[k]!=-1){
			node[k]=lazy[k];
			if(r-l>1){
				lazy[2*k]=lazy[k];
				lazy[2*k+1]=lazy[k];
			}
			lazy[k]=-1;
		}
	}
	
	void update(int a,int b,ll x,int k=1,int l=0,int r=-1){
		if(r<0) r=n;
		eval(k,l,r);
		if(r<=a||b<=l) return;
		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);
			node[k]=min(node[2*k],node[2*k+1]);
		}
	}
	
	ll getMin(int a,int b,int k=1,int l=0,int r=-1){
		if(r<0) r=n;
		eval(k,l,r);
		if(r<=a||b<=l) return INF;
		if(a<=l&&r<=b) return node[k];
		ll vl=getMin(a,b,2*k,l,(l+r)/2);
		ll vr=getMin(a,b,2*k+1,(l+r)/2,r);
		return min(vl,vr);
	}
};

int main(){
	int n,q; cin>>n>>q;
	LazySegmentTree tree(n);
	rep(i,q){
		int m,s,t,x;
		cin>>m;
		if(m==0){
			cin>>s>>t>>x;
			tree.update(s,t+1,x);
		}else{
			cin>>s>>t;
			cout<<tree.getMin(s,t+1)<<endl;
		}
	}
}