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;
typedef double db;
typedef pair<ll,ll> P;
#define pb push_back
#define ft first
#define sd second
#define mp make_pair
#define fr(i,n) for(int i=0;i<(n);i++)
#define Fr(i,n) for(int i=0;i++<(n);)
#define ifr(i,n) for(int i=(n)-1;i>=0;i--)
#define iFr(i,n) for(int i=(n);i>0;i--)

template<class T> struct LSMIN{
    int n;
    T e;
    vector<T> dat,lazy;
    vector<bool> vis;

    LSMIN(T n_,T e_){
        n=1;e=e_;
        while(n<n_) n*=2;
        dat.resize(2*n-1,e);
        lazy.resize(2*n-1,e);
        vis.resize(2*n-1);
    }

    void prop(int k,int l,int r){
        if(!vis[k]) return;
        dat[k]=lazy[k];
        if(r-l>1){
            lazy[2*k+1]=lazy[2*k+2]=lazy[k];
            vis[2*k+1]=vis[2*k+2]=1;
        }
        vis[k]=0;
        lazy[k]=e;
    }

    void ud(int be,int en,T t,int k,int l,int r){
        prop(k,l,r);
        if(r<=be||en<=l) return;
        if(be<=l&&r<=en){
            lazy[k]=t;
            vis[k]=1;
            prop(k,l,r);
            return;
        }
        ud(be,en,t,k*2+1,l,(l+r)/2);
        ud(be,en,t,k*2+2,(l+r)/2,r);
        dat[k]=min(dat[2*k+1],dat[2*k+2]);
    }

    void ud(int be,int en,T t){
        ud(be,en,t,0,0,n);
    }

    T gt(int be,int en,int k,int l,int r){
        if(r<=be||en<=l) return e;
        prop(k,l,r);
        if(be<=l&&r<=en) return dat[k];
        return min(gt(be,en,k*2+1,l,(l+r)/2),gt(be,en,k*2+2,(l+r)/2,r));
    }

    T gt(int be,int en){
        return gt(be,en,0,0,n);
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,q,s,t,x;
    bool f;
    cin>>n>>q;
    LSMIN<int> sg(n,INT_MAX);
    fr(i,q){
        cin>>f;
        if(f){
            cin>>s>>t;
            cout<<sg.gt(s,++t)<<endl;
        }
        else{
            cin>>s>>t>>x;
            sg.ud(s,++t,x);
        }
        //fr(j,sg.n*2-1) cout<<"["<<sg.dat[j]<<","<<sg.lazy[j]<<"] ";
        //cout<<endl;
    }
}