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 LL long long
#define maxn 200010
#define mod 1000000007
using namespace std;
int a[maxn];
LL ans=1,x,y;
int main(){
	int n,k,i,b,l,r;
	scanf("%d%d",&n,&k);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	sort(a+1,a+1+n);
	if(a[1]>=0){
		for(i=n;i>=n-k+1;i--)ans=ans*a[i]%mod;
	}else if(a[n]<0){
		if(k&1){
			for(i=n;i>=n-k+1;i--)ans=ans*a[i]%mod;
			ans=(ans%mod+mod)%mod;
		}else{
			for(i=1;i<=k;i++)ans=ans*a[i]%mod;
		}
	}else{
		int cnt1=0,cnt2=0;
		for(i=1;i<=n;i++)
			if(a[i]<0)cnt1++;
			else break;
		cnt2=n-cnt1;
		if(cnt1/2*2+cnt2<k){
			for(i=1;i<=n;i++)ans=ans*a[i]%mod;
			ans=(ans%mod+mod)%mod;
		}else{
			if(k&1)
				ans=a[n],n--,k--;
			l=1,r=n;
			while(k){
				x=(LL)a[l]*a[l+1],y=(LL)a[r]*a[r-1];
				if(x>y)x%=mod,ans=ans*x%mod,l+=2;
				else y%=mod,ans=ans*y%mod,r-=2;
				k-=2;
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}
