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<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

const int N=2E5+10,mod = 1000000007;
typedef long long LL;
int a[N];
LL res=1;

int main()
{
    int n,k;
    cin>>n>>k;

    for(int i=0;i<n;i++)cin>>a[i];

    sort(a,a+n);

    int sign=1;
    if(n==k){
        for(int i=0;i<n;i++)
            (res*=a[i])%=mod;
    }//n==k时
    else{
        int l=0,r=n-1;
        if(k%2)//奇数情况
        {
            res=a[r--],k--;
            if(res<0)sign=-1;//唯一的全负状况 取最小
        }
        while(k)//转化为偶数状态处理
        {
            LL ll = (LL)a[l]*a[l+1],rr=(LL)a[r]*a[r-1];
            if(ll*sign>rr*sign){
                res=ll%mod*res%mod;l+=2;
            }//左右两边哪个绝对值大就选哪个 全负数状态相反
            else {res=rr%mod*res%mod;r-=2;}
            k-=2;
        }
    }

    cout<<(res+mod)%mod;
    return 0;
}
