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;
typedef long long ll;
const ll maxn=2e5+100;
const ll inf=0x3f3f3f3f3f3f3f3f;
const ll mod=1e9+7;
ll t,n,k,a[maxn];
int main(){
    //freopen("in.txt","r",stdin);
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>n>>k;
    ll res=1,zheng=0,fu=0;
    for(int i=0;i<n;i++){
        cin>>a[i];
        if(a[i]>0) zheng++;
    }
    sort(a,a+n);
    if(zheng==0){
        if(k%2==1){
            for(int i=1;i<=k;i++){
                res=res*a[n-i]%mod;
            }
        }else{
            for(int i=0;i<k;i++){
                res=res*a[i]%mod;
            }
        }
        cout<<(res+mod)%mod<<endl;
        return 0;
    }
    if(k%2==0){
        zheng=n-1,fu=0;
    }else{
        zheng=n-2,fu=0,k-=2;//打的时候这里忘了减2了
        res=a[n-1];
    }
    for(;k>0&&zheng-1>=0&&fu+1<n;k-=2){
        if(a[zheng]*a[zheng-1]<a[fu]*a[fu+1]){
            res=res*a[fu]%mod*a[fu+1]%mod;
            fu+=2;
        }else{
            res=res*a[zheng]%mod*a[zheng-1]%mod;
            zheng-=2;
        }
    }
    while(res<0){
        res+=mod;
    }
    cout<<res<<endl;
    return 0;
}
