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 int long long
//BEGIN CUT HERE
struct RUP{
  int n;
  vector<int> dat,laz;
  const int def=INT_MAX;
  RUP(){}
  RUP(int n_){init(n_);}
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.clear();
    dat.resize(2*n-1,def);
    laz.clear();
    laz.resize(2*n-1,-1);
  }
  inline void eval(int len,int k){
    if(laz[k]<0) return;
    if(k*2+1<n*2-1){
      laz[k*2+1]=laz[k];
      laz[k*2+2]=laz[k];
    }
    dat[k]=laz[k];
    laz[k]=-1;
  }
  int update(int a,int b,int x,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return dat[k];
    if(a<=l&&r<=b) return laz[k]=x;
    eval(r-l,k);
    dat[k]=min(update(a,b,x,k*2+1,l,(l+r)/2),
	       update(a,b,x,k*2+2,(l+r)/2,r));
    return dat[k];
  }
  int query(int a,int b,int k,int l,int r){
    eval(r-l,k);
    if(r<=a||b<=l) return def;
    if(a<=l&&r<=b) return dat[k];
    int vl=query(a,b,k*2+1,l,(l+r)/2);
    int vr=query(a,b,k*2+2,(l+r)/2,r);
    return min(vl,vr);
  }
  void update(int a,int b,int x){
    update(a,b,x,0,0,n);
  }
  int query(int a,int b){
    return query(a,b,0,0,n);
  }
};
//END CUT HERE
signed main(){
  int n,q;
  cin>>n>>q;
  RUP rup(n);
  for(int i=0;i<q;i++){
    int f;
    cin>>f;
    if(!f){
      int s,t,x;
      cin>>s>>t>>x;
      rup.update(s,t+1,x);
    }else{
      int s,t;
      cin>>s>>t;
      cout<<rup.query(s,t+1)<<endl;
    }
  }
  return 0;
}

/*
verified on 2017/07/12
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_D
*/