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;
#define rep(i,n) for (int (i)=(0);(i)<(int)(n);++(i))
using ll = long long;
using namespace std;

struct LazySegmentTree {
private:
    int n;
    vector<ll> node, lazy;
    vector<bool> lazyflag;
    const ll inf;
public:
    LazySegmentTree(vector<ll> v) : inf((1LL<<31)-1) {
        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[2*i+1], node[2*i+2]);
    }

    // 伝搬させるため
    void eval(int k, int l, int r) {
        if (lazyflag[k]) {
            node[k] = lazy[k];

            // 範囲が半分になるから，値も半分になる
            if (r - l > 1) {
                lazy[2*k+1] = lazy[2*k+2] = lazy[k];
                lazyflag[2*k+1] = lazyflag[2*k+2] = true;
            }

            lazyflag[k] = false;
        }
    }

    // [a, b)
    void query(int a, int b, ll x, int k=0, int l=0, int r=-1) {
        if (r < 0) r = n;

        // k番目のノードで遅延評価
        eval(k, l, r);

        // 範囲外なら何もしない
        if (r <= a or b <= l) return;

        // 完全に被覆していたら評価
        if (a <= l and r <= b) {
            lazy[k] = x;
            lazyflag[k] = true;
            eval(k, l, r);
        }
        else {
            // そうじゃないなら，被覆するまで下がる
            query(a, b, x, 2*k+1, l, (l+r)/2);
            query(a, b, x, 2*k+2, (l+r)/2, r);
            node[k] = min(node[2*k+1], node[2*k+2]);
        }
    }

    ll getmin(int a, int b, int k=0, int l=0, int r=-1) {
        if (r < 0) r = n;
        if (r <= a or b <= l) return inf;

        eval(k, l, r);
        if (a <= l and r <= b) return node[k];
        ll vl = getmin(a, b, 2*k+1, l, (l+r)/2);
        ll vr = getmin(a, b, 2*k+2, (l+r)/2, r);
        return min(vl, vr);
    }
};

int main() {
    int n, q;
    cin >> n >> q;
    vector<ll> v(n, (1LL<<31)-1);

    LazySegmentTree lst(v);
    rep(i, q) {
        int c;
        cin >> c;
        if (c == 0) {
            ll s, t, x;
            cin >> s >> t >> x;
            lst.query(s, t+1, x);
        }
        else {
            ll s, t;
            cin >> s >> t;
            cout << lst.getmin(s, t+1) << endl;
        }
    }
}