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;
using ll = long long;

struct LazySegmentTree {
    int size;
    vector<int> node, lazy;
    const int I = (unsigned int) (1 << 31) - 1;

    LazySegmentTree(int n) {
        size = 1;
        while (size < n) size <<= 1;
        node.resize(2 * size, I);
        lazy.resize(2 * size, -1);
    }

    void eval(int k) {
        if (lazy[k] == -1) return;
        node[k] = lazy[k];
        if (k < size) lazy[2 * k] = lazy[2 * k + 1] = lazy[k];
        lazy[k] = -1;
    }

    void update(int a, int b, int x, int k = 1, int l = 0, int r = -1) {
        if (r == -1) r = size;
        eval(k);
        if (r <= a || b <= l) return;
        if (a <= l && r <= b) {
            lazy[k] = x;
            eval(k);
        } else {
            int m = (l + r) / 2;
            update(a, b, x, 2 * k, l, m);
            update(a, b, x, 2 * k + 1, m, r);
            node[k] = min(node[2 * k], node[2 * k + 1]);
        }
    }

    int query(int a, int b, int k = 1, int l = 0, int r = -1) {
        if (r == -1) r = size;

        eval(k);
        if (r <= a || b <= l) return I;
        if (a <= l && r <= b) return node[k];
        int m = (l + r) / 2;
        return min(query(a, b, 2 * k, l, m), query(a, b, 2 * k + 1, m, r));
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n, q;
    cin >> n >> q;
    LazySegmentTree st(n);
    for (int j = 0; j < q; j++) {
        int type;
        cin >> type;
        if (type == 0) {
            int s, t, x;
            cin >> s >> t >> x;
            st.update(s, t + 1, x);
        } else {
            int s, t;
            cin >> s >> t;
            cout << st.query(s, t + 1) << '\n';
        }
    }
}