#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using P = pair<ll, ll>;
using Graph = vector<vector<ll>>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int ddx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll MOD = 1000000007;
const ll INF = 1000000000000000000L;
#ifdef __DEBUG

/**
 * For DEBUG
 * https://github.com/ta7uw/cpp-pyprint
 */
#include "cpp-pyprint/pyprint.h"

#endif

void Main() {
    ll N;
    cin >> N;
    ll t = N / 25;
    ll ans = t;
    N -= 25 * t;
    t = N / 10;
    ans += t;
    N -= 10 * t;
    t = N / 5;
    ans += t;
    N -= 5 * t;
    ans += N;
    cout << ans << '\n';
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
    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;

using ll = long long;
using P = pair<ll, ll>;
using Graph = vector<vector<ll>>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int ddx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll MOD = 1000000007;
const ll INF = 1000000000000000000L;
#ifdef __DEBUG

/**
 * For DEBUG
 * https://github.com/ta7uw/cpp-pyprint
 */
#include "cpp-pyprint/pyprint.h"

#endif

void Main() {
    ll N;
    cin >> N;
    ll t = N / 25;
    ll ans = t;
    N -= 25 * t;
    t = N / 10;
    ans += t;
    N -= 10 * t;
    t = N / 5;
    ans += t;
    N -= 5 * t;
    ans += N;
    cout << ans << '\n';
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
    return 0;
}


```