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;
#define N (1<<18)

ll ai[N*2];
ll value[N*2];
ll max_num = 1;

void lazy_update(int k){
    if(value[k] == -1)return;
    if(k < N-1){
        value[k*2+1] = value[k];
        value[k*2+2] = value[k];
    }
    value[k] = -1;
}

ll update(int a, int b, int x, int k = 0, int l = 0, int r = N){
    if(value[k] != -1) ai[k] = value[k];
    if(r <= a || b <= l) return ai[k];
    if(a <= l && r <= b){
        value[k] = ai[k] = x;
        return ai[k];
    }
    lazy_update(k);
    ll vl = update(a, b, x, k*2+1, l, (l+r)/2);
    ll vr = update(a, b, x, k*2+2, (l+r)/2, r);
    ai[k] = min(vl, vr);
    return ai[k];
}

ll find(int a, int b, int k = 0, int l = 0, int r = N){
    if(value[k] != -1) ai[k] = value[k];
    if(r <= a || b <= l) return max_num;
    if(a <= l && r <= b) return ai[k];
    lazy_update(k);
    ll resultl = find(a, b, k*2+1, l, (l+r)/2);
    ll resultr = find(a, b, k*2+2, (l+r)/2, r);
    return min(resultl, resultr);
}

int main(){
    for(int i = 0; i < 31; i++) max_num *= 2;
    max_num -= 1;

    for(int i = 0; i < N*2; i++){
        ai[i] = max_num;
        value[i] = -1;
    }

    int n, q;
    cin >> n >> q;

    for(int i = 0; i < q; i++){
        int com, s, t, x;
        cin >> com;
        if(com == 0){
            cin >> s >> t >> x;
            update(s, t+1, x);
        }else{
            cin >> s >> t;
            cout << find(s, t+1) << endl;
        }
    }
}