#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    ll ans = 0;
    int num[] = {25, 10, 5, 1};
    for(int i = 0; i < 4; i++)
    {
        ans += n / num[i];
        n %= num[i];
    }
    cout << ans << endl;
    return 0;
}



### Prompt

Please create a solution in Cpp to the following problem:

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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    ll ans = 0;
    int num[] = {25, 10, 5, 1};
    for(int i = 0; i < 4; i++)
    {
        ans += n / num[i];
        n %= num[i];
    }
    cout << ans << endl;
    return 0;
}



```