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;

struct LazySegmentTree{
    private:
        int n;
        vector<int> node, lazy;
        vector<bool> lazyFlag;
    
    public:
        LazySegmentTree(vector<int> v){
            int sz = (int)v.size();
            n = 1; while(n < sz) n*= 2;
            node.resize(2 * n - 1);
            lazy.resize(2 * n - 1, INT_MAX);
            lazyFlag.resize(2 * n - 1, false);

            for(int i = 0; i < sz; i++) node[i+n-1] = v[i];
            for(int i = n - 2; i >= 0; i--) node[i] = min(node[i*2+1], node[i*2+2]);
        }

        void lazyEvaluate(int k, int l, int r){
            if(lazyFlag[k]){
                node[k] = lazy[k];
                if(r - l > 1){
                    lazy[k*2+1] = lazy[k*2+2] = lazy[k];
                    lazyFlag[k*2+1] = lazyFlag[k*2+2] = true;
                }
                lazyFlag[k] = false;
            }
        }
        void update(int a, int b, int x, int k = 0, int l = 0, int r = -1){
            if(r < 0) r = n;
            lazyEvaluate(k, l, r);
            if(b <= l || r <= a) return;
            if(a <= l && r <= b){
                lazy[k] = x;
                lazyFlag[k] = true;
                lazyEvaluate(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);
                node[k] = min(node[2*k+1], node[2*k+2]);
            }
        }

        int find(int a, int b, int k = 0, int l = 0, int r = -1){
            if(r < 0) r = n;
            lazyEvaluate(k, l, r);
            if(b <= l || r <= a) return INT_MAX;
            if(a <= l && r <= b) return node[k];
            int vl = find(a, b, 2 * k + 1, l, (l + r) / 2);
            int vr = find(a, b, 2 * k + 2, (l + r) / 2, r);
            return min(vl, vr);
        }
};

int main(){
    int n, q; cin >> n >> q;
    LazySegmentTree seg(vector<int>(n, INT_MAX));
    while(q--){
        int query; cin >> query;
        if(query == 0){
            int s,t,x; cin >> s >> t >> x;
            seg.update(s, t + 1, x);
        }else{
            int s,t; cin >> s >> t;
            cout << seg.find(s, t + 1) << endl;
        }
    }
}