#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i, a, n) for (int i = (a); i < (n); i++)
#define repr(i, n, a) for (int i = (n)-1; i >= (a); i--)
#define each(it, v) for (auto(it) = (v).begin(); (it) != (v).end(); (it)++)
#define all(v) (v).begin(), (v).end()
#define vi vector<int>
#define vvi vector<vi>
#define vd vector<double>
#define vvd vector<vd>
#define co(x) cout << (x)
#define con(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define out(x) rep(i, 0, (x).size()) cout << (x)[i] << ((i != (x).size() - 1) ? " " : "\n")
const double PI = 3.14159265;
const int MAXN = 1 << 15;
const int LINF = 1LL << 60;
const int INF = 1 << 30;
#define Pr(a) (#a) << "=" << (a) << " "
#define pr(a) cout << Pr(a) << '\n'
///////////////////////////////////////////////////////

signed main() {
  int n;
  cin >> n;
  int ans = 0;
  int a[4] = {25, 10, 5, 1};
  for (int i = 0; i < 4; i++) {
    ans += n / a[i];
    n %= a[i];
  }
  cout << ans << endl;
  return 0;
}


### Prompt

In CPP, your task is to solve 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>
using namespace std;
#define int long long
#define rep(i, a, n) for (int i = (a); i < (n); i++)
#define repr(i, n, a) for (int i = (n)-1; i >= (a); i--)
#define each(it, v) for (auto(it) = (v).begin(); (it) != (v).end(); (it)++)
#define all(v) (v).begin(), (v).end()
#define vi vector<int>
#define vvi vector<vi>
#define vd vector<double>
#define vvd vector<vd>
#define co(x) cout << (x)
#define con(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define out(x) rep(i, 0, (x).size()) cout << (x)[i] << ((i != (x).size() - 1) ? " " : "\n")
const double PI = 3.14159265;
const int MAXN = 1 << 15;
const int LINF = 1LL << 60;
const int INF = 1 << 30;
#define Pr(a) (#a) << "=" << (a) << " "
#define pr(a) cout << Pr(a) << '\n'
///////////////////////////////////////////////////////

signed main() {
  int n;
  cin >> n;
  int ans = 0;
  int a[4] = {25, 10, 5, 1};
  for (int i = 0; i < 4; i++) {
    ans += n / a[i];
    n %= a[i];
  }
  cout << ans << endl;
  return 0;
}


```