#include <bits/stdc++.h>
    using namespace std;
    #define ll long long
    #define MOD 1000000007
    #define print(A,n) for(ll i=0;i<n;++i)cout<<A[i]<<' ';cout<<endl;
    #define take(A,n) for(ll i=0;i<n;++i)cin>>A[i];
    //    for(auto& it : m)
    void fastio(){ios_base::sync_with_stdio(false);cin.tie(NULL);}
    //  priority_queue<ll int> pq;
    //  priority_queue <ll int, vector<ll int>, greater<ll int> > pq;
    //  vector<ll int> v;
    //  vector< pair<ll int,ll int> > v;
    //  set<ll int>::iterator it=s.begin();
    //  unordered_map<  pair<ll int,ll int>, ll int> mp;
     
     
     
    ll int power(ll int x, ll int y, ll int p)  
    {  
        ll int res = 1;     // Initialize result  
      
        x = x % p; // Update x if it is more than or  
                    // equal to p 
       
        if (x == 0) return 0; // In case x is divisible by p; 
      
        while (y > 0)  
        {  
            // If y is odd, multiply x with result  
            if (y & 1)  
                res = (res*x) % p;  
      
            // y must be even now  
            y = y>>1; // y = y/2  
            x = (x*x) % p;  
        }  
        return res;  
    }  
     
    void computeLPSArray(string pat, int M, int* lps) 
    { 
        // length of the previous longest prefix suffix 
        int len = 0; 
      
        lps[0] = 0; // lps[0] is always 0 
      
        // the loop calculates lps[i] for i = 1 to M-1 
        int i = 1; 
        while (i < M) { 
            if (pat[i] == pat[len]) { 
                len++; 
                lps[i] = len; 
                i++; 
            } 
            else // (pat[i] != pat[len]) 
            { 
                // This is tricky. Consider the example. 
                // AAACAAAA and i = 7. The idea is similar 
                // to search step. 
                if (len != 0) { 
                    len = lps[len - 1]; 
      
                    // Also, note that we do not increment 
                    // i here 
                } 
                else // if (len == 0) 
                { 
                    lps[i] = 0; 
                    i++; 
                } 
            } 
        } 
    }
    bool subset[10005][1005]; 
    void isSubsetSum(ll int set[],ll int n,ll int sum) 
{ 
    // The value of subset[i][j] will be true if 
    // there is a subset of set[0..j-1] with sum 
    // equal to i 
    
  
    // If sum is 0, then answer is true 
    
  
    // If sum is not 0 and set is empty, 
    // then answer is false 
    
  for (int i = 0; i <= n; i++) 
        subset[i][0] = true; for (int i = 1; i <= sum; i++) 
        subset[0][i] = false; 
    // Fill the subset table in botton up manner 
    for (int i = 1; i <= n; i++) { 
        for (int j = 1; j <= sum; j++) { 
            if (j < set[i - 1]) 
                subset[i][j] = subset[i - 1][j]; 
            if (j >= set[i - 1]) 
                subset[i][j] = subset[i - 1][j] 
                               || subset[i - 1][j - set[i - 1]]; 
        } 
    } 
  
    /*   // uncomment this code to print table 
     for (int i = 0; i <= n; i++) 
     { 
       for (int j = 0; j <= sum; j++) 
          printf ("%4d", subset[i][j]); 
       printf("\n"); 
     }*/
  
   
} 
ll int maxProductSubarrayOfSizeK(ll int A[],ll int n, ll int k) 
{ 
    // sorting given input array 
    sort(A, A + n); 
  
    // variable to store final product of all element 
    // of sub-sequence of size k 
    ll int product = 1; 
  
    // CASE I 
    // If max element is 0 and 
    // k is odd then max product will be 0 
    if (A[n - 1] == 0 && (k & 1)) 
        return 0; 
  
    // CASE II 
    // If all elements are negative and 
    // k is odd then max product will be 
    // product of rightmost-subarray of size k 
    if (A[n - 1] <= 0 && (k & 1)) { 
        for (ll int i = n - 1; i >= n - k; i--) 
            {product *= A[i]; 
                product+=MOD;
                    product%=MOD;}
        return product; 
    } 
  
    // else 
    // i is current left pointer index 
    ll  int i = 0; 
  
    // j is current right pointer index 
    ll int j = n - 1; 
  
    // CASE III 
    // if k is odd and rightmost element in 
    // sorted array is positive then it 
    // must come in subsequence 
    // Multiplying A[j] with product and 
    // correspondingly changing j 
    if (k & 1) { 
        product *= A[j]; 
          product+=MOD;
        product%=MOD;
        j--; 
        k--; 
    } 
  
    // CASE IV 
    // Now k is even 
    // Now we deal with pairs 
    // Each time a pair is multiplied to product 
    // ie.. two elements are added to subsequence each time 
    // Effectively k becomes half 
    // Hence, k >>= 1 means k /= 2 
    k >>= 1; 
  
    // Now finding k corresponding pairs 
    // to get maximum possible value of product 
    for (ll int itr = 0; itr < k; itr++) { 
  
        // product from left pointers 
        ll int left_product = A[i] * A[i + 1]; 
        
  
        // product from right pointers 
       ll int right_product = A[j] * A[j - 1]; 
         
  
        // Taking the max product from two choices 
        // Correspondingly changing the pointer's position 
        if (left_product > right_product) { 
              left_product+=MOD;
        left_product%=MOD;
            product *= left_product; 
              product+=MOD;
            product%=MOD;
            i += 2; 
        } 
        else { 
            right_product+=MOD;
        right_product%= MOD; 
            product *= right_product; 
              product+=MOD;
            product%=MOD;
            j -= 2; 
        } 
    } 
  
    // Finally return product 
    return product; 
} 
     
    int main()
    {
      fastio();
      
    ll int T;
    T=1;
   
 
    while(T--)
    { 
 
        ll int n,m,k,i;
       cin>>n>>k;
       ll int a[n];
       for(i=0;i<n;i++)
      {  cin>>a[i];
          }cout<<maxProductSubarrayOfSizeK(a,n,k);
   }
     
        
      
     
     
     
        return 0;
    }

Measured memory_bytes: 5104