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>
using namespace std;

const int INF = (1<<31)-1;
const int MAX = (1<<17)-1;
int n, D[1000000], lazy[1000000];
bool flag[1000000];
void init(int n_ ){
    n = 1;
    while(n<n_) n *= 2;
    for(int i=0; i < 2*n-1; i++){
        D[i] = ((int)1<<31)-1;
        flag[i] = false;
    }
}
void eval(int k, int l, int r){
    if(flag[k]){
      D[k] = lazy[k];
        if(l+1 < r){
          lazy[2*k+1] = lazy[k];
          lazy[2*k+2] = lazy[k];
          flag[2*k+1] = true;
          flag[2*k+2] = true;
        }
        flag[k] = false;
    }
}
void update(int a,int b,int x,int k,int l,int r){
    eval(k, l, r);
    if(r <= a || b <= l)return;
    if(a <= l && r <= b){
        flag[k] = true;
        lazy[k] = x;
        eval(k, l, r);
    }
    else{
        update(a, b, x, 2*k+1, l, (l+r)/2);
        update(a, b, x, 2*k+2, (l+r)/2, r);
        D[k]=min(D[2*k+1],D[2*k+2]);
    }
}
void update(int a, int b, int x) { update(a, b, x, 0, 0, n);}

int query(int a, int b, int k, int l, int r){
    eval(k, l, r);
    if(r <= a|| b <= l) return INF;
    if(a <= l && r <= b) return D[k];
    int vl=query(a, b, 2*k+1, l, (l+r)/2);
    int vr=query(a, b, 2*k+2, (l+r)/2 ,r);
    return min(vl,vr);
}
int findmin(int a, int b) { return query(a, b, 0, 0, n); }


int main(void){
    int p, q;

    cin >> p >> q;
    init(p);

    for(int i=0; i<q; i++){
        int com, x, y, z, m;
        cin >> com;

        if(com == 0){
           cin >> x >> y >> z;
            update(x, y+1, z);
        }else{
          cin >> x >> y;
            cout << findmin(x, y+1) << endl;
        }
    }
    return 0;
}