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.*;

public class d{
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt(), K = in.nextInt();
		int[][] pairs = new int[N][];
		int sumFives = 0;
		for(int n=0;n<N;n++){
			long a = in.nextLong();
			pairs[n] = new int[]{0,0};
			for(; a % 2 == 0; a /= 2){
				pairs[n][0]++;
			}
			for(; a % 5 == 0; a /= 5){
				pairs[n][1]++;
			}
			sumFives += pairs[n][1];
		}
		
		int[][] dp = new int[sumFives+1][K+1];
		for(int[] a : dp) Arrays.fill(a, -1);
		dp[0][0] = 0;
		
		int ans = 0;
		for(int n=0;n<N;n++){
			int d2 = pairs[n][0], d5 = pairs[n][1];
			for(int s = sumFives; s >= d5; s--){
				for(int k = K; k > 0; k--){
					if(dp[s-d5][k-1] != -1){
						dp[s][k] = Math.max(dp[s][k], dp[s-d5][k-1] + d2);
					}
				}
			}
		}

		for(int s=0;s<=sumFives;s++){
//			System.out.println(Arrays.toString(dp[s]));
			ans = Math.max(ans, Math.min(s, dp[s][K]));
		}
		
		System.out.println(ans);
	}
}