Given are N integers A_1,\ldots,A_N.

We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.

Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Constraints

* 1 \leq K \leq N \leq 2\times 10^5
* |A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:


N K
A_1 \ldots A_N


Output

Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Examples

Input

4 2
1 2 -3 -4


Output

12


Input

4 3
-1 -2 -3 -4


Output

1000000001


Input

2 1
-1 1000000000


Output

1000000000


Input

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1


Output

999983200
#include <bits/stdc++.h>

using namespace std;

#define intt long long
#define FAST ios_base::sync_with_stdio(0); cin.tie(0);

const int MOD = 1e9 + 7;

int n, k;

int main() {
	
	FAST;
	
	cin >> n >> k;
	vector <int> a(n);
	for (int i = 0; i < n; i ++) {
		cin >> a[i];
	}
	sort(a.begin(), a.end());
	if (k % 2 && a[n - 1] == 0) {
		cout << 0 << '\n';
		return 0;
	}
	int ans = 1;
	if (k % 2 && a[n - 1] < 0) {
		for (int i = 1; i <= k; i ++) {
			ans = 1LL * ans * a[n - i] % MOD;
			if (ans < 0) ans += MOD;
		}
		cout << ans << '\n';
		return 0;
	}
	int i = 0, j = n - 1;
	if (k % 2 && a[n - 1] > 0) {
			ans = a[n - 1];
			j --;
			k --;
	}
	k /= 2;
	while (k -- ) {
		intt left = 1LL * a[i] * a[i + 1];
		intt right = 1LL * a[j] * a[j - 1];
		if (left > right) {
			ans = 1LL * ans * (left % MOD) % MOD;
			if (ans < 0) ans += MOD;
			i += 2;
		}
		else {
			ans = 1LL * ans * (right % MOD) % MOD;
			if (ans < 0) ans += MOD;
			j -= 2;
		}
	}
	cout << ans << '\n';
	return 0;
}