#include <bits/stdc++.h>

int main()
{
    long long N;
    int cents[] = {25, 10, 5, 1};
    std::scanf("%d", &N);
    long long sum = 0;
    for (int i = 0; i < 4; i++) {
        if (N == 0) break;
        sum += int(N / cents[i]);
        N = N % cents[i];
    }
    std::printf("%d\n", sum);
    return 0;
}


### Prompt

Develop a solution in Cpp to the problem described below:

You want to make change for $ n $ cents. Assuming that you have infinite supply of coins of 1, 5, 10 and / or 25 cents coins respectively, find the minimum number of coins you need.

Constraints

* $ 1 \ le n \ le 10 ^ 9 $

Input


$ n $


The integer $ n $ is given in a line.

output

Print the minimum number of coins you need in a line.

Examples

Input

100


Output

4


Input

54321


Output

2175

### Response

```cpp
#include <bits/stdc++.h>

int main()
{
    long long N;
    int cents[] = {25, 10, 5, 1};
    std::scanf("%d", &N);
    long long sum = 0;
    for (int i = 0; i < 4; i++) {
        if (N == 0) break;
        sum += int(N / cents[i]);
        N = N % cents[i];
    }
    std::printf("%d\n", sum);
    return 0;
}


```