#include "bits/stdc++.h"
#define LF '\n'
#define ALL(x) x.begin(), x.end()
#define iostreamBooster() { cin.tie(nullptr); ios_base::sync_with_stdio(false); }
using namespace std;
typedef int64_t i64;
typedef pair<int,int> pii;
template<class A, class B>inline bool chmax(A &a, const B &b){return b>a ? a=b,1 : 0;}
template<class A, class B>inline bool chmin(A &a, const B &b){return b<a ? a=b,1 : 0;}
constexpr int INF = 0x3f3f3f3f;
template<class Itr> void dump(Itr begin, Itr end) { for (;begin != end; ++begin) clog << *begin << LF; }

signed main()
{
  int n;
  cin >> n;

  int ans = 0;
  for (int e : {25, 10, 5, 1}) {
    ans += n / e;
    n -= e * (n/e);
  }

  cout << ans << LF;

  return 0;
}



### Prompt

Please provide a cpp coded solution 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"
#define LF '\n'
#define ALL(x) x.begin(), x.end()
#define iostreamBooster() { cin.tie(nullptr); ios_base::sync_with_stdio(false); }
using namespace std;
typedef int64_t i64;
typedef pair<int,int> pii;
template<class A, class B>inline bool chmax(A &a, const B &b){return b>a ? a=b,1 : 0;}
template<class A, class B>inline bool chmin(A &a, const B &b){return b<a ? a=b,1 : 0;}
constexpr int INF = 0x3f3f3f3f;
template<class Itr> void dump(Itr begin, Itr end) { for (;begin != end; ++begin) clog << *begin << LF; }

signed main()
{
  int n;
  cin >> n;

  int ans = 0;
  for (int e : {25, 10, 5, 1}) {
    ans += n / e;
    n -= e * (n/e);
  }

  cout << ans << LF;

  return 0;
}



```