#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<ld> vd;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<pair<ll, ll>> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define FOR(i, a, b) for (ll i = (ll)a; i < (ll)b; i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); i++)
#define FI first
#define SE second
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define TS to_string
#define NP next_permutation
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) (ll) x.size()
#define SP(x) setprecision((ll)x)


ll INF = 1e9;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "Yes" << endl;}
void no(){ cout << "No" << endl;}

//-----------------------------------------
//-----------------------------------------

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  ll n;
  cin >> n;
  ll ans = 0;
  ans += n / 25;
  n %= 25;
  ans += n / 10;
  n %= 10;
  ans += n / 5;
  n %= 5;
  ans += n;
  cout << ans <<endl;

  return 0;
}



### Prompt

Construct a CPP code solution to the problem outlined:

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;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<ld> vd;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<pair<ll, ll>> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define FOR(i, a, b) for (ll i = (ll)a; i < (ll)b; i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); i++)
#define FI first
#define SE second
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define TS to_string
#define NP next_permutation
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) (ll) x.size()
#define SP(x) setprecision((ll)x)


ll INF = 1e9;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "Yes" << endl;}
void no(){ cout << "No" << endl;}

//-----------------------------------------
//-----------------------------------------

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  ll n;
  cin >> n;
  ll ans = 0;
  ans += n / 25;
  n %= 25;
  ans += n / 10;
  n %= 10;
  ans += n / 5;
  n %= 5;
  ans += n;
  cout << ans <<endl;

  return 0;
}



```