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;

struct dat{
  bool delay; // ?????¶????????????1
  int dvalue; // ?????¶?????????
  int min_value; // ???????????????????°????
  dat(){};
  dat(bool a,int b,int c){
    delay=a,dvalue=b,min_value=c;
  };
};

const int MAX_N=(1<<17);

dat d[2*MAX_N];

void init(){
  for(int i=0;i<2*MAX_N;i++) d[i]=dat(0,0,INT_MAX);
}

void Delay(int k,int l,int r){
  if(r-l==1||d[k].delay==0)return ;
  d[k].delay=0;
  d[k*2+1]=d[k*2+2]=dat(1,d[k].dvalue,d[k].dvalue);
}

void Range_Update(int a,int b,int x,int k=0,int l=0,int r=MAX_N){
  Delay(k,l,r);
  if(r<=a||b<=l)return ;
  if(a<=l&&r<=b){
    d[k].delay=true;
    d[k].dvalue=x;
    d[k].min_value=x;
    return ;
  }
  Range_Update(a,b,x,k*2+1,(l+r)/2,r);
  Range_Update(a,b,x,k*2+2,l,(l+r)/2);
  d[k].min_value=min(d[k*2+1].min_value,d[k*2+2].min_value);
}

int Range_Minimum(int a,int b,int k=0,int l=0,int r=MAX_N){
  Delay(k,l,r);
  if(r<=a||b<=l)return INT_MAX;
  if(a<=l&&r<=b)return d[k].min_value;
  int res=INT_MAX;
  res=min(res,Range_Minimum(a,b,k*2+1,(l+r)/2,r));
  res=min(res,Range_Minimum(a,b,k*2+2,l,(l+r)/2));
  return res;
}

int main(){
  
  init();

  int n,q;

  cin>>n>>q;

  int a,l,r,x;
  for(int i=0;i<q;i++){
    cin>>a>>l>>r;
    
    if(!a){
      
      cin>>x;
      Range_Update(l,r+1,x);
      
    }else cout<<Range_Minimum(l,r+1)<<endl;

  }
  
  return 0;
}