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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static long MOD = 1000000007;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String[] s = br.readLine().split(" ");

        int n = Integer.parseInt(s[0]);
        int k = Integer.parseInt(s[1]);

        String[] s2 = br.readLine().split(" ");
        Integer[] nums = new Integer[n];
        List<Long> positives = new ArrayList<>();
        List<Long> negatives = new ArrayList<>();


        for (int i = 0; i < n; i++) {
            nums[i] = (Integer.parseInt(s2[i]));
            int temp = Integer.parseInt(s2[i]);
            if (temp<0)negatives.add((long) (temp));
            else  positives.add((long) temp);
        }

        boolean isPositive = false;

        int posCount = Math.min(k,positives.size());
        int negCount = k - posCount;

        if (negCount%2==0)isPositive = true;
        else{
            if(posCount-1>=0 && negCount+1<=negatives.size())isPositive=true;
        }

        if (isPositive){
            //positive sing = > select negatives or positives in groups of two (by whichever pair has higher product) and mul 1 more pos at the end if k nos not taken

            positives.sort(Collections.reverseOrder());
            Collections.sort(negatives);

            int negIdx = 0;
            int posIdx = 0;
            int total = k;
            long ans  = 1L;
            if(k%2==1){
                ans = positives.get(posIdx);
                ++posIdx;
                --total;
            }
            while (total>0){

                long negProd = negatives.size()-negIdx>=2 ? (negatives.get(negIdx)*negatives.get(negIdx+1)) : 0;
                long posProd = positives.size()-posIdx>=2 ? (positives.get(posIdx)*positives.get(posIdx+1)) : 0;


                ans = (ans * (Math.max(negProd,posProd)%MOD))%MOD;
                if(negProd>posProd)
                    negIdx += 2;
                else
                    posIdx += 2;
                total -= 2;
            }
            System.out.println(ans);
        }else{
            // sign will be negative, hence choose smallest k numbers(by abs value).
            Arrays.sort(nums, Comparator.comparingInt(Math::abs));
            long ans = 1L;
            for(int i = 0 ; i<k; i++){
                ans = (ans*nums[i])%MOD;
            }
            System.out.println((ans+MOD)%MOD);
        }

    }
}
