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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define endl '\n'
#define ld long double
#define all(a) a.begin(),a.end()
#define int long long
#define pb push_back
#define pii pair <int, int>
#define ff first
#define ss second
#define sz(v) (int)v.size()
#define UB upper_bound
#define LB lower_bound
const int INF = 1e18 + 0;
const int mod = 1e9 + 7;
//const int mod = 998244353;

int mul(int a, int b) {
  a += mod; b += mod;
  a %= mod; b %= mod; return ((a * b) % mod);
}

void solve() {
  int n, k, ans = 1, K; cin >> n >> k;
  K = k;
  vector <int> p, neg, z;
  vector <pii> v;
  for(int i = 0; i < n; i++) {
    int x, s; cin >> x;
    if(x > 0) {
      p.pb(x);
      s = 1;
    }
    if(x < 0) {
      neg.pb(abs(x));
      s = -1;
    }
    if(x == 0) {
      z.pb(x);
      s = 1;
    }
    v.pb({abs(x), s});
  }
  sort(all(p)); sort(all(neg)); sort(all(z)); sort(all(v));
  int i = sz(p)-1, j = sz(neg)-1;

  if(k%2) {
    if(!p.empty()) {
      k--; ans = mul(ans, p[i]);
      i--;
    }
  }

  while(k && (i > 0 || j > 0)) {
     int a = -INF, b = -INF;
     if(i >= 1) {
       a = p[i] * p[i-1];
     }
     if(j >= 1) {
       b = neg[j] * neg[j-1];
     }
     if(a > b) {
       ans = mul(ans, a);
       k-=2; i-=2;
     }
     else {
       ans = mul(ans, b);
       k-=2; j-=2;
     }
  }
  if(k) {
    if(!z.empty()) {
      ans = 0;
    } else {
      ans = 1;
      for(int l = 0; l < K; l++) {
         ans = mul(ans, v[l].ss * v[l].ff);
      }
    }
  }
  cout << ans << '\n';
}

signed main() {
  ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

  #ifndef ONLINE_JUDGE
   freopen("input.txt", "r", stdin);
   freopen("output.txt", "w", stdout);
  #endif
     solve();
  }