Codeforces-837D：Round Subset（DP）

D. Round Subset

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

3 250 4 20

output

3

input

5 315 16 3 25 9

output

3

input

3 39 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.

思路：因为末尾0的个数来源于2和5的个数。所以用d[i][j][k]表示前i个数中取j个数且因子5的个数为k时，因子2的个数的最大值。ans=max（d[i][j][k]，k）。

#include<bits/stdc++.h>using namespace std;int d[2][210][6000],n,m;int five(__int64 x){    int sum=0;    while(x%5==0)sum++,x/=5;    return sum;}int two(__int64 x){    int sum=0;    while(x%2==0)sum++,x/=2;    return sum;}int main(){    __int64 x;    int ans=0,pre=0,now=1;    memset(d,-1,sizeof d);    d[0][0][0]=0;    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    {        scanf("%I64d",&x);        int f=five(x),t=two(x);        for(int j=0;j<=i&&j<=m;j++)        {            for(int k=0;k<=26*i;k++)            {                if(j+1<=i&&j+1<=m&&d[pre][j][k]!=-1)d[now][j+1][k+f]=max(d[now][j+1][k+f],d[pre][j][k]+t);                d[now][j][k]=max(d[now][j][k],d[pre][j][k]);                ans=max(ans,min(d[now][m][k],k));            }        }        memset(d[pre],-1,sizeof d[pre]);//二维滚动数组        pre^=1;        now^=1;    }    cout<<ans<<endl;    return 0;}