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;
const ll MAX_N = 1<< 20 ;
typedef pair<ll,ll> P;
ll n;
P dat[2*MAX_N-1];
 

void init(ll a){
  n = 1;
  while( n < a ) { n *=2;}
  for(ll i=0;i<2*n-1;i++) { dat[i]=P(INT_MAX,INT_MAX);}
}
 
ll update(ll a,ll b, ll  k, ll l, ll r,ll x){
  if(r<=a||b<=l)return dat[k].second;
  if(a<=l&&r<=b){
    dat[k]=P(x,x);
    return x;
  }else{
    if(dat[k].first!=INT_MAX){
      dat[k*2+1].first=dat[k*2+1].second=dat[k].first;
      dat[k*2+2].first=dat[k*2+2].second=dat[k].first;
      dat[k].first=INT_MAX;
    }
    ll m=(l+r)/2;
    ll v1=update(a,b,k*2+1,l,m,x);
    ll v2=update(a,b,k*2+2,m,r,x);
    return dat[k].second=min(v1,v2);
  }
}
 
ll query(ll a,ll b,ll k,ll l,ll r){
  if(r<=a||b<=l)return INT_MAX;
  if(a<=l&&r<=b)return dat[k].second;
  else{
    if(dat[k].first!=INT_MAX){
      dat[k*2+1].first=dat[k*2+1].second=dat[k].first;
      dat[k*2+2].first=dat[k*2+2].second=dat[k].first;
      dat[k].first=INT_MAX;
    }
    return min(query(a,b,k*2+1,l,(l+r)/2),query(a,b,k*2+2,(l+r)/2,r));
  }
}

int main(){
  ll N,Q; cin >> N >> Q;
  init(N);
  for(ll i = 0; i < Q; i++){
    int com; cin >> com;
    if( com ){
      ll s,t; cin >> s >> t; cout << query(s,t+1,0,0,n) << endl;}
    else{
      ll s,t,x; cin >> s >> t >> x; update(s,t+1,0,0,n,x) ;}
  }
  
  return 0;}