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;
int dp[2][201][200 * 25 + 1], t[2][201][200 * 25 + 1];
int main() {
  int n, k;
  cin >> n >> k;
  for (int x = 0; x < 2; ++x)
    for (int i = 0; i <= n; ++i)
      for (int j = 0; j < 200 * 25 + 1; ++j) dp[x][i][j] = INT_MIN;
  dp[0][0][0] = dp[1][0][0] = 0;
  for (int tx = 1; tx <= n; ++tx) {
    long long a;
    int x = 0, y = 0;
    for (int x = 0; x < 2; ++x)
      for (int i = 0; i <= n; ++i)
        for (int j = 0; j < 200 * 25 + 1; ++j) t[x][i][j] = INT_MIN;
    cin >> a;
    while (a % 2 == 0) ++x, a /= 2;
    while (a % 5 == 0) ++y, a /= 5;
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < 200 * 25 + 1; ++j) {
        if (dp[0][i][j] == INT_MIN) continue;
        int c = min(j + x, y);
        int nx = min(200 * 25 + 1 - 1, j + x - c),
            ny = min(200 * 25 + 1 - 1, y - c);
        if (nx == 0) t[1][i + 1][ny] = max(t[1][i + 1][ny], dp[0][i][j] + c);
        if (ny == 0) t[0][i + 1][nx] = max(t[0][i + 1][nx], dp[0][i][j] + c);
      }
    }
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < 200 * 25 + 1; ++j) {
        if (dp[1][i][j] == INT_MIN) continue;
        int c = min(x, j + y);
        int nx = min(200 * 25 + 1 - 1, x - c),
            ny = min(200 * 25 + 1 - 1, j + y - c);
        if (nx == 0) t[1][i + 1][ny] = max(t[1][i + 1][ny], dp[1][i][j] + c);
        if (ny == 0) t[0][i + 1][nx] = max(t[0][i + 1][nx], dp[1][i][j] + c);
      }
    }
    for (int x = 0; x < 2; ++x)
      for (int i = 0; i <= k; ++i)
        for (int j = 0; j < 200 * 25 + 1; ++j)
          dp[x][i][j] = max(dp[x][i][j], t[x][i][j]);
  }
  int ans = 0;
  for (int x = 0; x < 2; ++x)
    for (int j = 0; j < 200 * 25 + 1; ++j) ans = max(ans, dp[x][k][j]);
  cout << ans << '\n';
  return 0;
}
