Educational Round 26 D. Round Subset

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, ..., a_n (1 ≤ a_i ≤ 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.

每输入一个数，就把分解为x个2，和y个5，则就转化为背包问题

#include <iostream>
#include <stdio.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int dp[205][8010];
int tdp[205][8010];
int in[205][2];
int main() {
    int N, K, i, j, k;
    scanf("%d %d", &N, &K);
    for (i = 1; i <= N; i++) {
        ll t;
        scanf("%lld", &t);
        while (t % 2 == 0) {
            in[i][0]++;
            t /= 2;
        }
        while (t % 5 == 0) {
            in[i][1]++;
            t /= 5;
        }
    }

    for (i = 0; i <= K; i++)
        for (j = 0; j <= 8000; j++) dp[i][j] = -INF;
            dp[0][0] = 0;
    for (i = 1; i <= N; i++) {
        for (j = 0; j <= K; j++)
            for (k = 0; k <= 8000; k++)
                tdp[j][k] = dp[j][k];
        for (j = 0; j < K; j++) {
            for (k = 0; k <= 8000; k++) {
                int j2 = j + 1, k2 = k + in[i][1];
                if (k2 <= 8000) tdp[j2][k2] = max(tdp[j2][k2], dp[j][k] + in[i][0]);
            }
        }
        for (j = 0; j <= K; j++) for (k = 0; k <= 8000; k++) dp[j][k] = tdp[j][k];
    }

    int ans = 0;
    for (i = 0; i <= K; i++) for (j = 0; j <= 8000; j++) ans = max(ans, min(j, dp[i][j]));
    return !printf("%d
", ans);
}