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; 
int main(){
 ios_base::sync_with_stdio(false);
   cin.tie(NULL);
   int n,m,k;
   cin>>n>>k;
   vector<long long int> A(n);
   for(int i=0;i<n;i++){
   	cin>>A[i];
   }
  sort(A.begin(),A.end());
  long long int mod=1000000007; 
   long long int ans=1;
    if (A[n - 1] == 0 && (k & 1)) 
        {
        	cout<<"0\n";
        	return 0;
		}
    if (A[n - 1] <= 0 && (k & 1)) { 
        for (int i = n - 1; i >= n - k; i--) 
            ans=((ans%mod)*(A[i]%mod))%mod;
            cout<<(ans+mod)<<"\n";
            return 0;
    } 
    int i = 0; 
    int j = n - 1; 
    if (k & 1) { 
       ans=((ans%mod)*(A[j]%mod))%mod;
        j--; 
        k--; 
    } 
    k >>= 1; 
    for (int itr = 0; itr < k; itr++) { 
        long long int left_product =A[i]*A[i+1]; 
        long long int right_product = A[j]*A[j-1]; 
        if (left_product > right_product) { 
            ans=((ans%mod)*(left_product%mod))%mod; 
            i += 2; 
        } 
        else { 
            ans=((ans%mod)*( right_product%mod))%mod;
            j -= 2; 
        } 
    } 
    ans=(ans+mod)%mod;
    cout<<ans<<"\n"; 
   return 0;
}