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 pair<int,int> P;
typedef long long ll;
const int MAX_N = 1<<17;
const ll INF = INT_MAX;

class RMQ2{
public : 
  struct data{
    bool type; //0 - empty   , 1 - update
    ll value;
  };
  
  ll n,dat[2*MAX_N-1];
  data td[2*MAX_N-1];
  
  RMQ2(){
    n=MAX_N;
    memset(td,0,sizeof(td));
    for(int i=0;i<2*n-1;i++)dat[i]=INF;
  }
  RMQ2(int n_){
    n=1;
    while(n<n_)n*=2;
    memset(td,0,sizeof(td));
    for(int i=0;i<2*n-1;i++)dat[i]=INF;
  }
  
  ll update(int a,int b,int x,int k=0,int l=0,int r=-1){
    if(r==-1) r=n;
    if(r<=a||b<=l)return dat[k];
    if(a<=l&&r<=b){
      td[k]=(data){1,x};
      return dat[k]=x;
    }
    
    if(td[k].type){
      dat[k*2+1]=dat[k*2+2]=td[k].value;
      td[k*2+1]=td[k*2+2]=(data){1,td[k].value};
    }
    td[k].type=0;
    
    ll vl=update(a,b,x,k*2+1,l,(l+r)/2);
    ll vr=update(a,b,x,k*2+2,(l+r)/2,r);
    return dat[k]=min(vl,vr);
  }
  
  ll find(int a,int b,int k=0,int l=0,int r=-1){
    if(r==-1) r=n;
    if(r<=a||b<=l)return INT_MAX;
    if(a<=l&&r<=b)return dat[k];
    
    if(td[k].type){
      dat[k*2+1]=dat[k*2+2]=td[k].value;
      td[k*2+1]=td[k*2+2]=(data){1,td[k].value};
    }
    td[k].type=0;
    ll vl=find(a,b,k*2+1,l,(l+r)/2);
    ll vr=find(a,b,k*2+2,(l+r)/2,r);
    return min(vl,vr);
  }
};
 
int main(){
  int q,n;
  cin>>n>>q;
  RMQ2 T(n);
  while(q--){
    int cmd;
    cin>>cmd;
    if(cmd==0){
      int s,t,x;
      cin>>s>>t>>x;
      T.update(s,t+1,x);
    }
    else {
      int s,t;
      cin>>s>>t;
      cout <<T.find(s,t+1)<<endl;
    }
  }
  return 0;
}