#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector> 
#include<limits>
#include<numeric>
#include<math.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i = (int)(a); i < (int)(b); i++)
#define in(x) cin >> x
#define out(str) cout << str << endl

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int N,ans=0;

int main(){
    cin>>N;
    ans+=N/25;
    N%=25;
    ans+=N/10;
    N%=10;
    ans+=N/5;
    N%=5;
    ans+=N;
    out(ans);
}


### 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<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector> 
#include<limits>
#include<numeric>
#include<math.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i = (int)(a); i < (int)(b); i++)
#define in(x) cin >> x
#define out(str) cout << str << endl

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int N,ans=0;

int main(){
    cin>>N;
    ans+=N/25;
    N%=25;
    ans+=N/10;
    N%=10;
    ans+=N/5;
    N%=5;
    ans+=N;
    out(ans);
}


```