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>
#define len(v) ((int)((v).size()))
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define chmax(x, v) x = max((x), (v))
#define chmin(x, v) x = min((x), (v))
using namespace std;
using ll = long long;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	const int MOD = 1e9+7;
	int n,k;
	cin >> n >> k;
	vector<ll> neg,po;
	for (int i = 0; i < n; ++i) {
		ll x; cin >> x;
		if (x > 0) po.push_back(x);
		else neg.push_back(x);
	}
	sort(rall(po)); // 9 8 7 ...
	sort(all(neg)); // -9 -8 ...
	int i1 = min(k,len(po));
	int i2 = k-i1;
	if (i2 % 2 == 1) { 
		if (i1 == 0 || i2 == len(neg)) {
			for (int x : neg) po.push_back(-x);
			sort(all(po));
			ll res = (MOD-1);
			for (int i = 0; i < k; ++i) {
				res = (res*po[i])%MOD;
			}
			cout << res << "\n";
			return 0;
		}
		--i1; ++i2;
	}
	while (i1 >= 2 && i2+1 < len(neg) && po[i1-1]*po[i1-2] < neg[i2]*neg[i2+1]) {
		i1 -= 2;
		i2 += 2;
	}
	ll res = 1;
	for (int i = 0; i < i1; ++i) {
		res = (res*po[i])%MOD;
	}
	for (int i = 0; i < i2; ++i) {
		res = (res*(neg[i]+MOD))%MOD;
	}
	cout << res << "\n";
}
