Problem:
Score : 500 points

Problem Statement

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).

Sample Input 1

4 2
1 2 -3 -4

Sample Output 1

12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

Sample Input 2

4 3
-1 -2 -3 -4

Sample Output 2

1000000001

The possible products of the three chosen elements are -24, -12, -8, and -6, so the maximum product is -6.

We print this value modulo (10^9+7), that is, 1000000001.

Sample Input 3

2 1
-1 1000000000

Sample Output 3

1000000000

The possible products of the one chosen element are -1 and 1000000000, so the maximum product is 1000000000.

Sample Input 4

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1

Sample Output 4

999983200

Be sure to print the product modulo (10^9+7).

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; ++i)
using namespace std;

long gyaku(long a, long mod){
  long s = 1, tmp = a, t = mod - 2;
  while(t > 0){
    if(t % 2 == 1){
      s *= tmp;
      s %= mod;
    }
    t /= 2;
    tmp *= tmp;
    tmp %= mod;
  }
  return s;
}

int main(){
  long ans = 1, mod = 1000000007; int n, l = 0, la = 0, lb = 0, k; cin >> n >> k;
  vector<long> v(n), a = {}, b = {}; rep(i, n) cin >> v[i];
  sort(v.begin(), v.end());
  while(l < n){
    if(l < n-1 && v[l] < 0 && v[l+1] < 0){
      b.push_back(v[l] * v[l+1]);
      l += 2;
    }
    else{
      a.push_back(v[l]);
      ++l;
    }
  }
  reverse(a.begin(), a.end());
  while(k > 0){
    if(k > 1){
      if(la < (int)a.size() && lb <(int) b.size()){
        if(a[la] * a[la + 1] > b[lb]){
          ans *= a[la];
          ans %= mod;
          ++la;
          --k;
        }
        else{
          ans *= b[lb] % mod;
          ans %= mod;
          ++lb;
          k -= 2;
        }
      }
      else if(la < (int)a.size()){
        ans *= a[la];
        ans %= mod;
        ++la;
        --k;
      }
      else{
        ans *= b[lb] % mod;
        ans %= mod;
        ++lb;
        k -= 2;
      }
    }
    else{
      if(lb < (int)b.size() && b[lb] > a[la-1] * a[la]){
        ans *= gyaku(a[la-1], mod);
        ans %= mod;
        ans *= b[lb] % mod;
        ans %= mod;
        --k;
      }
      else{
        ans *= a[la];
        ans %= mod;
        --k;
      }
    }
  }
  cout << ans << "\n";
  return 0;
}