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>
#define int long long
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

template<typename T> class LazySegmentTree
{
 private:
    int n;
    vector<T> node,lazy;
 public:
    LazySegmentTree(int sz,T init){
        n=1;
        while(n<sz) n*=2;
        node.assign(2*n-1,init);
        lazy.assign(2*n-1,-1);
    }
    void update(int a,int b,int x,int k,int l,int r){
        if(a<=l and r<=b) node[k]=lazy[k]=x;
        else{
            if(lazy[k]!=-1) push(k),lazy[k]=-1;
            if(check(a,b,l,(l+r)/2)) update(a,b,x,2*k+1,l,(l+r)/2);
            if(check(a,b,(l+r)/2,r)) update(a,b,x,2*k+2,(l+r)/2,r);
            node[k]=min(node[2*k+1],node[2*k+2]);
        }
    }
    T get(int a,int b,int k,int l,int r){
        if(a<=l and r<=b) return node[k];
        else{
            if(lazy[k]!=-1) push(k),lazy[k]=-1;
            T res=INT_MAX;
            if(check(a,b,l,(l+r)/2)) res=min(res,get(a,b,2*k+1,l,(l+r)/2));
            if(check(a,b,(l+r)/2,r)) res=min(res,get(a,b,2*k+2,(l+r)/2,r));
            return res;
        }
    }
    void push(int k){
        node[2*k+1]=node[2*k+2]=lazy[k];
        lazy[2*k+1]=lazy[2*k+2]=lazy[k];
    }
    bool check(int a,int b,int l,int r){
        return max(a,l)<min(b,r);
    }
    void update(int a,int b,int x){ update(a,b,x,0,0,n); }
    T get(int a,int b){ return get(a,b,0,0,n); }
};

signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N,Q; cin>>N>>Q;
    LazySegmentTree<int> seg(N,INT_MAX);
    while(Q--){
        int q; cin>>q;
        if(q==0){
            int s,t,x; cin>>s>>t>>x;
            seg.update(s,t+1,x);
        }else{
            int s,t; cin>>s>>t;
            cout<<seg.get(s,t+1)<<endl;
        }
    }
    return 0;
}