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;
}