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:
//update(s,t,x): a[s],a[s+1],...,a[t] をxに変更する
//find(s,t): a[s],a[s+1],...,a[t]の最小値を出力する

#include<bits/stdc++.h>
using namespace std;

#define int long long
const int sqrtN = 512;
const int INF = (1LL << 31) - 1;

struct SqrtDecomposition{
    int N, K;
    vector<int> data;
    vector<int> bucketMin;
    vector<bool> lazyFlag;
    vector<int> lazyUpdate;
    SqrtDecomposition(int n):N(n){
        K = (N + sqrtN - 1) / sqrtN;
        data.assign(K * sqrtN, INF);
        bucketMin.assign(K, INF);
        lazyFlag.assign(K, false);
        lazyUpdate.assign(K, 0);
    }

    void eval(int k){
        if(lazyFlag[k]){
            lazyFlag[k] = false;
            for(int i = k * sqrtN; i < (k + 1) * sqrtN; i++){
                data[i] = lazyUpdate[k];
            }
        }
    }

    void update(int s, int t, int x){
        for(int k = 0; k < K; k++){
            int l = k * sqrtN;
            int r = (k + 1) * sqrtN;
            if(r <= s || t <= l) continue;

            if(s <= l && r <= t){
                bucketMin[k] = x;
                lazyFlag[k] = true;
                lazyUpdate[k] = x;
            }else{
                eval(k);
                for(int i = max(s, l); i < min(t, r); i++){
                    data[i] = x;
                }

                bucketMin[k] = INF;
                for(int i = l; i < r; ++i){
                    bucketMin[k] = min(bucketMin[k], data[i]);
                }
            }
        }
    }

    //[s, t)
    int find(int s, int t){
        int minVal = INF;
        for(int k = 0; k < K; k++){
            int l = k * sqrtN;
            int r = (k + 1) * sqrtN;
            if(r <= s || t <= l) continue;

            if(s <= l && r <= t){
                minVal = min(minVal, bucketMin[k]);
            }else{
                eval(k);
                for(int i = max(s, l); i < min(t, r); i++){
                    minVal = min(minVal, data[i]);
                }
            }
        }

        return minVal;
    }
};

signed main() {
    int N, Q; cin >> N >> Q;
    SqrtDecomposition decomp(N);
    while(Q--){
        int c; cin >> c;
        if(c == 0){
            int s, t, x; cin >> s >> t >> x;
            decomp.update(s, t + 1, x);
        }else{
            int s, t; cin >> s >> t;
            cout << decomp.find(s, t + 1) << endl;
        }
    }
    return 0;
}