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;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
const int INF=1e9+5;
const int N=2e5+5;
const ll mod=1e9+7;

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	ll n,k;
	cin>>n>>k;
	vector<ll> a(n);
	vector<ll> pos,neg;
	int zeros=0;
	for(ll& x:a){
		cin>>x;
		if(x>0) pos.push_back(x);
		if(x<0) neg.push_back(abs(x));
		if(!x) ++zeros;
	}
	int ps=(int)pos.size(),ns=(int)neg.size();
	if((ps+ns/2*2>=k)&&(ps>0||k%2==0)){
		sort(rall(pos));
		sort(rall(neg));
		ll start=1;
		if(k&1) start=pos[0];
		vector<ll> ar;
		for(int i=k%2;i<ps-1;i+=2) ar.push_back(pos[i]*pos[i+1]);
		for(int i=0;i<ns-1;i+=2) ar.push_back(neg[i]*neg[i+1]);
		sort(rall(ar));	
		for(int i=0;i<k/2;++i){
			(start*=ar[i]%mod)%=mod;
		}
		cout<<start%mod<<endl;
	}
	else{
		for(int i=0;i<n;++i) a[i]=abs(a[i]);
		sort(all(a));
		ll ans=1;
		for(int i=0;i<k;++i) (ans*=a[i]%mod)%=mod;
		cout<<(mod-ans)%mod<<endl;
	}
	return 0;
}
