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
import java.util.*;

public class Main {
    private final static long MOD = 1000000007;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        Long ans = 1L;

        Integer[] a = new Integer[n];
        LinkedList<Integer> s = new LinkedList<>();
        LinkedList<Integer> t = new LinkedList<>();

        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            if (a[i] < 0) t.add(a[i]);
            else s.add(a[i]);
        }

        int S = s.size(), T = t.size();
        boolean ok = false;

        if (S > 0) {
            if (n == k) ok = (T % 2 == 0);
            else ok = true;
        } else ok = (k % 2 == 0);

        if (!ok) {
            Arrays.sort(a, (x, y) -> (Math.abs(x) - Math.abs(y)));
            for (int i = 0; i < k; i++) {
                ans = mul(ans, a[i]);
            }
        } else {
            Collections.sort(s);
            Collections.sort(t, (x, y) -> (y - x));
            if (k % 2 == 1) {
                ans = mul(ans, s.pollLast());
            }
            ArrayList<Long> p = new ArrayList<>();
            while (s.size() >= 2) {
                long x = s.pollLast();
                x *= s.pollLast();
                p.add(x);
            }
            while (t.size() >= 2) {
                long x = t.pollLast();
                x *= t.pollLast();
                p.add(x);
            }
            Collections.sort(p, (o1, o2) -> Long.compare(o2, o1));
            for (int i = 0; i < k / 2; i++) {
                ans = mul(ans, p.get(i));
            }
        }
        ans = Math.floorMod(ans, MOD);
        System.out.println(ans);

    }

    private static long mul(long a, long b) {
        return (((a + MOD) % MOD) * ((b + MOD) % MOD)) % MOD;
    }
}