Codeforces 837D

D. Round Subset

time limit per test           2 seconds

memory limit per test     256 megabytes

Let's call theroundness of the number the number of zeros to which it ends.

You have an array ofn numbers. You need to choose a subsetof 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 numbersn and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).

The second line containsn space-separated integer numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10¹⁸).

Output

Print maximal roundness of product of the chosen subset of lengthk.

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 are3 subsets of 2 numbers. [50, 4] has product 200 with roundness2,[4, 20] — product 80,roundness1,[50, 20] — product 1000,roundness3.

In the second example subset[15, 16, 25] has product 6000,roundness3.

In the third example all subsets has product withroundness0.

 

【题意】

给出n个数，让你从中选出k个数，使得它们乘积的后缀0最多，输出这个值。

【思路】

显然0是由2*5得来的，所以先算出一个数分解质因子后2和5的个数。

然后用dp[i][j]表示选择了i个数，且这i个数质因子中2的个数和为j时,质因子中5的个数和的最大值。状态转移方程为：

dp[j][l]=max(dp[j][l],dp[j-1][l-n2]+n5);

最后扫一遍即可(注意取2的个数与5的个数其中的较小值)

#include <cstdio>#include <algorithm>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define rush() int T;scanf("%d",&T);while(T--)typedef long long ll;const int maxn = 205;const int maxm = 205*64;const ll mod = 1e9+7;const int INF = 0x3f3f3f;const double eps = 1e-9;int dp[maxn][maxm];int n,k;ll x;int main(){    while(~scanf("%d%d",&n,&k))    {        for(int i=0;i<=k;i++)        for(int j=0;j<maxm;j++)        {            dp[i][j]=-INF;        }        dp[0][0]=0;        for(int i=0;i<n;i++)        {            int n2=0,n5=0;            scanf("%I64d",&x);            while(x%2==0) x/=2,n2++;            while(x%5==0) x/=5,n5++;            for(int j=k;j>=1;j--)            for(int l=n2;l<maxm;l++)            {                dp[j][l]=max(dp[j][l],dp[j-1][l-n2]+n5);            }        }        int ans=0;        for(int i=1;i<maxm;i++)        {            ans=max(ans,min(i,dp[k][i]));        }        printf("%d\n",ans);    }    return 0;}