Let's call the roundness of the number the number of zeros to which it ends.

You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.

Input

The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).

The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).

Output

Print maximal roundness of product of the chosen subset of length k.

Examples

Input

3 2
50 4 20


Output

3


Input

5 3
15 16 3 25 9


Output

3


Input

3 3
9 77 13


Output

0

Note

In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.

In the second example subset [15, 16, 25] has product 6000, roundness 3.

In the third example all subsets has product with roundness 0.
import java.io.*; import java.util.*;
public class CF0837D {

    /**
     * n at most 200 inspires an O(n^3) solution
     * Let dp[i][j][k]=In [1,i], pick j #s st sum of v5 is k, find max sum of v2
     */
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int k=Integer.parseInt(st.nextToken());
        st=new StringTokenizer(br.readLine());
        int[] v2=new int[n]; int[] v5=new int[n];
        int cnt; long a;
        for (int i = 0; i < n; i++) {
            cnt=0; a=Long.parseLong(st.nextToken());
            while(a%2==0){
                a/=2; cnt++;
            }
            v2[i]=cnt; cnt=0;
            while(a%5==0){
                a/=5; cnt++;
            }
            v5[i]=cnt; 
        }
        
        int[][] dp=new int[k+1][k*26];//Need memory compression
        for (int i = 0; i <=k; i++) {
            for (int j = 0; j < k*26; j++) {
                dp[i][j]=-1;//-1 means unreachable
            }
        }
        dp[0][0]=0;
        int dp2[][]=new int[k+1][k*26];
        int best=0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= Math.min(k,i); j++) {
                for (int l = 0; l < k*26; l++) {
                    dp2[j][l]=dp[j][l];//Discard
                    if(j>0 && l>=v5[i-1] && dp[j-1][l-v5[i-1]]>=0){
                        dp2[j][l]=Math.max(dp[j][l],dp[j-1][l-v5[i-1]]+v2[i-1]);//Keep
                        
                    }
                    if(i==n && j==k){
                        best=Math.max(best,Math.min(l,dp2[j][l]));
                        
                    }
                }
            }
            for (int j = Math.min(k,i)+1; j <=k; j++) {
                for (int l = 0; l <k*26; l++) {
                    dp2[j][l]=-1;
                }
            }
            for (int j = 0; j <=k; j++) {
                for (int l = 0; l <k*26; l++) {
                    dp[j][l]=dp2[j][l];
                }
            }
        }
        
        System.out.println(best);
    }
    
}
