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 ALL(A) (A).begin(), (A).end()
#define ll long long
#define rep(i, n) for (int i = 0; i < (n); i++)

using namespace std;

const ll mod = 1e9 + 7;

int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  vector<int> s, t;
  rep(i,n) {
    cin >> a[i];
    if (a[i] < 0) t.push_back(a[i]);
    else s.push_back(a[i]);
  }
 
  int S = s.size();
  int T = t.size();
  bool ok = false;
  if (S > 0) {
    if (n == k) ok = (T%2==0);
    else ok = true;
  } else {
    ok = (k%2==0);
  }
 
  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 {
    sort(s.begin(), s.end());
    sort(t.rbegin(), t.rend());
    if (k%2 == 1) {
      ans *= s.back();
      s.pop_back();
    }
    vector<ll> p;
    while (s.size() >= 2) {
      ll x = s.back(); s.pop_back();
      x *= s.back(); s.pop_back();
      p.push_back(x);
    }
    while (t.size() >= 2) {
      ll x = t.back(); t.pop_back();
      x *= t.back(); t.pop_back();
      p.push_back(x);
    }
    sort(p.rbegin(), p.rend());
    rep(i,k/2) ans *= p[i]%mod,ans %= mod;
  }
 
  cout << (ans + mod) % mod << endl;
  return 0;
}
