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 <iostream>
#include <algorithm>
using namespace std;
#define INF 2147483647
typedef long long ll;
int n,q;
ll seg[1<<19],lazy[1<<19];
void rep(int k,int r,int l){
    if(lazy[k]!=-1){
        seg[k]=lazy[k];
        if(l-r>0){
            lazy[k*2+1]=lazy[k];
            lazy[k*2+2]=lazy[k];
        }
    }
    lazy[k]=-1;
}
void update(int a,int b,int r,int l,int k,ll x){
    rep(k,r,l);
    if(a<=r&&l<=b){
        lazy[k]=x;
        rep(k,r,l);
        return;
    }
    if(l<a||b<r)return;
    update(a,b,r,(r+l-1)/2,k*2+1,x);
    update(a,b,(r+l+1)/2,l,k*2+2,x);
    seg[k]=min(seg[k*2+1],seg[k*2+2]);
    
}
ll find(int a,int b,int r,int l,int k){
    rep(k,r,l);
    if(a<=r&&l<=b)return seg[k];
    if(l<a||b<r)return INF;
    return min(find(a,b,r,(r+l-1)/2,k*2+1),find(a,b,(r+l+1)/2,l,k*2+2));
}
int main(void){
    cin>>n>>q;
    for(int i=0;i<1<<19;i++){
        seg[i]=INF;
        lazy[i]=-1;
    }
    for(int i=0;i<q;i++){
        int a,s,t;
        cin>>a>>s>>t;
        if(a){
            cout<<find(s,t,0,(1<<18)-1,0)<<endl;
        }else{
            ll x;
            cin>>x;
            update(s,t,0,(1<<18)-1,0,x);
        }
    }
    
}