#include<bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i) 
#define PI acos(-1)

using namespace std;
using ll = long long;
using P = pair<int, int>;
using LP = pair<ll, ll>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin>>n;

    vector<int> coins {25, 10, 5, 1};
    int cnt = 0;
    for (auto c : coins) {
        cnt += n / c;
        n %= c;
    }
    cout<<cnt<<endl;;
}

### 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<bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i) 
#define PI acos(-1)

using namespace std;
using ll = long long;
using P = pair<int, int>;
using LP = pair<ll, ll>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin>>n;

    vector<int> coins {25, 10, 5, 1};
    int cnt = 0;
    for (auto c : coins) {
        cnt += n / c;
        n %= c;
    }
    cout<<cnt<<endl;;
}

```