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;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define ll long long int
#define pb push_back
#define mp make_pair
const ll mod=1e9+7;

ll func(ll A[], ll n, ll k)
{
    sort(A, A + n);
    ll product = 1;
    if (A[n - 1] == 0 && (k & 1))
        return 0;
    if (A[n - 1] <= 0 && (k & 1)) {
        for (int i = n - 1; i >= n - k; i--)
            product=((product%mod)*(A[i]%mod))%mod;
        return (product+mod)%mod;
    }
    ll i = 0;
    ll j = n - 1;
  if (k & 1) {
        product *= A[j];
        j--;
        k--;
    }
        k >>= 1;
    for (int itr = 0; itr < k; itr++) {
        ll left_product = A[i] * A[i + 1];
        ll right_product = A[j] * A[j - 1];
        if (left_product > right_product) {
            product=((product%mod)*(left_product%mod))%mod;
            i += 2;
        }
        else {
            product=((product%mod)*(right_product%mod))%mod;
            j -= 2;
        }
    }
    return (product+mod)%mod;
}

void solve()
{
  ll n,k;
  cin>>n>>k;
  ll V[n];
  for(int i=0;i<n;i++)
  cin>>V[i];
  cout<<(func(V,n,k)+mod)%mod;
}

int main()
{
  fastio
  solve();
}
