https://codeforces.com/problemset/problem/837/D
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, roundness3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
/*
*@Author: STZG
*@Language: C++
*/
//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
using namespace std;
typedef long long ll;
const int N=200+10;
const int M=N*64;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f;
int t,n,m,k;
ll a[N];
int cnt2[N],cnt5[N];
int dp[N][N*64];
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++){
scanf("%I64d",&a[i]);
cnt2[i]=0;
cnt5[i]=0;
ll tmp=a[i];
while(tmp%2==0){
cnt2[i]++;
tmp/=2;
}
//tmp=a[i];
while(tmp%5==0){
cnt5[i]++;
tmp/=5;
}
}
for(int i=0;i<N;i++){
for(int j=0;j<M;j++){
dp[i][j]=-INF;
}
}
dp[0][0]=0;
int ans=0;
for(int i=1;i<=n;i++){
for(int j=k;j>=1;j--){
for(int l=cnt2[i];l<M;l++){
dp[j][l]=max(dp[j][l],dp[j-1][l-cnt2[i]]+cnt5[i]);
}
}
}
for(int i=1;i<M;i++){
ans=max(ans,min(i,dp[k][i]));
}
cout << ans << endl;
//cout << "Hello world!" << endl;
return 0;
}