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.
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 100000;
int dp[205][8010];
int tdp[205][8010];
int in[205][2];
int main() {
  int N, K, i, j, k;
  scanf("%d%d", &N, &K);
  for (i = 1; i <= N; i++) {
    long long t;
    scanf("%lld", &t);
    while (t % 2 == 0) {
      in[i][0]++;
      t /= 2;
    }
    while (t % 5 == 0) {
      in[i][1]++;
      t /= 5;
    }
  }
  for (int i = 0; i <= K; i++) {
    for (int j = 0; j <= 8000; j++) {
      dp[i][j] = -inf;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j <= K; j++) {
      for (int k = 0; k <= 8000; k++) {
        tdp[j][k] = dp[j][k];
      }
    }
    for (int j = 0; j < K; j++) {
      for (int k = 0; k <= 8000; k++) {
        int j2 = j + 1, k2 = k + in[i][1];
        if (k2 <= 8000) tdp[j2][k2] = max(tdp[j2][k2], dp[j][k] + in[i][0]);
      }
    }
    for (int j = 0; j <= K; j++) {
      for (int k = 0; k <= 8000; k++) {
        dp[j][k] = tdp[j][k];
      }
    }
  }
  int ans = 0;
  for (int i = 0; i <= K; i++)
    for (int j = 0; j <= 8000; j++) ans = max(ans, min(j, dp[i][j]));
  return !printf("%d\n", ans);
}
