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 = 0x3fffffff;
const int maxn = 205;
const int maxm = 205 * 64;
int dp[maxn][maxm];
int n, k;
long long in[maxn];
int a[maxn];
int b[maxn];
void fun(int i, long long num) {
  int cnt = 0;
  while (num % 2 == 0) {
    num /= 2;
    cnt++;
  }
  a[i] = cnt;
  cnt = 0;
  while (num % 5 == 0) {
    num /= 5;
    cnt++;
  }
  b[i] = cnt;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> in[i];
    fun(i, in[i]);
  }
  for (int i = 0; i <= k; i++) {
    for (int j = 1; j <= maxm; j++) {
      dp[i][j] = -INF;
    }
  }
  dp[0][0] = 0;
  long long x;
  for (int i = 0; i < n; i++) {
    for (int j = k; j >= 1; j--)
      for (int l = a[i]; l < maxm; l++) {
        dp[j][l] = max(dp[j][l], dp[j - 1][l - a[i]] + b[i]);
      }
  }
  int ans = 0;
  for (int i = 0; i <= maxm; i++) {
    ans = max(ans, min(i, dp[k][i]));
  }
  cout << ans << endl;
  return 0;
}
