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.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Tomas {

	static int n;
	static int k;
	
	static int[] pow5;
	static int[] pow2;
	
	static int ans = 0;
	
	public static void main(String[] args) {
		
		Scanner in = new Scanner(System.in);
		
		n = in.nextInt();
		k = in.nextInt();
		
		pow5 = new int[n];
		pow2 = new int[n];
		
		for(int i=0; i<n; i++) {
		
			long num = in.nextLong();
			
			pow5[i] = (get(num, 5));
			pow2[i] = (get(num, 2));
		
		}
		
		int[][][] dp = new int[2][k+1][5201];
		
		for(int i=0; i<1; i++) {
			for(int j=0; j<=k; j++) {
				Arrays.fill(dp[i][j], -99999999);
			}
		}
		
		int cur = 0;
		
		dp[cur][0][0] = 0;
		dp[cur][1][pow5[0]] = pow2[0];
		
		ans = Math.min(pow5[0], pow2[0]);
				
		for(int i=1; i<n; i++) {
			
			cur = 1 - cur;
			
			for(int j=0; j<=k; j++) {
				
				for(int f=0; f<5201; f++) {
		
					int pick = -99999999;
					int no = -99999999;
					
					if(j-1 >= 0 && f-pow5[i] >= 0)
						pick = pow2[i] + dp[1 - cur][j - 1][f - pow5[i]];
					
					no = dp[1 - cur][j][f];
					
					dp[cur][j][f] = Math.max(pick, no);
					
					ans = Math.max(ans, Math.min(f, dp[cur][j][f]));

				}
			}
		}
	
		System.out.println(ans);
	}


	private static int get(long num, int i) {
		
		int ans = 0;
		
		while(num % i == 0) {
			num /= i;
			ans++;
		}
		
		return ans;
	}

}
