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 be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=(1LL<<31)-1;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;
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, INF);
        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]);
        }
    }

    ll 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 INF;
        if(a <= l && r <= b) return node[k];
        ll vl = find(a, b, 2*k+1, l, (l+r)/2);
        ll vr = find(a, b, 2*k+2, (l+r)/2, r);
        return min(vl, vr);
    }
};
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    int n,q;
    cin >> n >> q;
    LazySegmentTree seg(vector<int> (n,INF));
    while(q--){
    	int a,b,c,d;
    	cin>>a>>b>>c;
    	if(a){
    		cout << seg.find(b,c+1) << endl;
    	}else{
    		cin>>d;
    		seg.update(b,c+1,d);
    	}
    }
    return 0;
}