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>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll=long long;

const int mod=1e9+7;

int main(){
    int n,k;
    cin>>n>>k;
    vector<int>a(n),s,t;
    rep(i,n){
        cin>>a[i];
        if(a[i]>=0)s.push_back(a[i]);
        else t.push_back(a[i]);
    }
    bool ok=false;
    if(!s.empty()){
        if(n==k){
            if(t.size()%2==0)ok=true;
        }
        else ok=true;
    }
    else if(k%2==0)ok=true;
    ll ans=1;
    if(!ok){
        sort(a.begin(),a.end(),[](int x,int y){
            return abs(x)<abs(y);
        });
        rep(i,k){
            ans*=a[i];
            ans%=mod;
        }
    }
    else{
        priority_queue<ll>q;
        sort(s.rbegin(),s.rend());
        sort(t.begin(),t.end());
        if(t.size()>=1)for(int i=0; i<t.size()-1; i+=2){
            q.push((ll)t[i]*t[i+1]);
        }
        int i=0;
        if(k%2==1)ans*=s[i],i++,ans%=mod;
        if(s.size()>=1)for(; i<s.size()-1; i+=2){
            q.push((ll)s[i]*s[i+1]);
        }
        rep(i,k/2){
            ll x=q.top()%mod;
            ans*=x;
            ans%=mod;
            q.pop();
        }
    }
    if(ans<0)ans+=mod;
    cout<<ans<<endl;
}