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 N = 201, M = 25, T = N * M;
int dp[2][N][T];
int two[N], five[N];
void solve() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    long long x;
    scanf("%lld", &x);
    while (x % 2 == 0) {
      two[i]++;
      x /= 2;
    }
    while (x % 5 == 0) {
      five[i]++;
      x /= 5;
    }
  }
  int maxi = 0;
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j <= k; ++j) {
      for (int z = 0; z < T; ++z) {
        dp[i][j][z] = -1e9;
      }
    }
  }
  dp[0][0][0] = 0;
  int pos = 0;
  for (int i = 0; i < n; ++i) {
    int u = min(i, k);
    int npos = pos ^ 1;
    int v = min(maxi + 30, T);
    for (int j = 0; j <= u + 1; ++j) {
      for (int z = 0; z < v; ++z) {
        dp[npos][j][z] = -1e9;
      }
    }
    int tmp = maxi;
    for (int j = 0; j <= u; ++j) {
      for (int z = maxi; z >= 0; --z) {
        dp[npos][j + 1][z + five[i]] =
            max(dp[npos][j + 1][z + five[i]], dp[pos][j][z] + two[i]);
        dp[npos][j][z] = max(dp[npos][j][z], dp[pos][j][z]);
        if (dp[npos][j + 1][z + five[i]] >= 0) {
          tmp = max(tmp, z + five[i]);
        }
      }
    }
    maxi = tmp;
    pos = npos;
  }
  int ret = 0;
  for (int i = 0; i < T; ++i) {
    int u = min(i, dp[pos][k][i]);
    if (dp[pos][k][i] >= 0) {
    }
    ret = max(ret, u);
  }
  printf("%d\n", ret);
}
int main() { solve(); }
