Educational Codeforces Round 26 D 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 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.

题意:给你n个数 取出k个 ans 为k个数乘积的结果的末尾的零的个数

题解:dp[i][j] 选择i个数  因子5的个数为j  的2的个数为 dp[i][j]

#pragma comment(linker, "/STACK:102400000,102400000")
#include <bits/stdc++.h>
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <string>
#include <complex>
#define LL long long
#define mod 1000000007
using namespace std;
int n,k;
LL a[205];
LL dp[205][7000];
struct node{
    int x,y;
}N[205];
int main()
{
    memset(dp,0,sizeof(dp));
    scanf("%d %d",&n,&k);
    for(int i=1; i<=n; i++)
        scanf("%I64d",&a[i]);
    for(int i=0;i<=k;i++)
        for(int e=0;e<=6998;e++)
         dp[i][e]=-1e9;
    dp[0][0]=0;
    LL zha;
    int now=0,xx=0;
    for(int i=1; i<=n; i++){
        zha=a[i];
        now=0;
        xx=0;
        while(zha>0){
            if(zha%5!=0)
                break;
            zha/=5;
            now++;
        }
        zha=a[i];
        while(zha>0){
            if(zha%2!=0)
                break;
            zha/=2;
            xx++;
        }
        N[i].x=now;
        N[i].y=xx;
    }
    for(int i=1;i<=n;i++){
        for(int j=k-1;j>=0;j--){
             for(int e=0;e<=6998;e++)
             dp[j+1][e+N[i].x]=max(dp[j+1][e+N[i].x],dp[j][e]+N[i].y);
        }
    }
    LL maxn=0;
    for(LL e=0;e<=6998;e++)
        maxn=max(maxn,min(dp[k][e],e));
    printf("%I64d
",maxn);
    return 0;
}