#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <stdio.h>
#include <bitset>
#include <complex>
#include <iomanip>
using namespace std;
#define MM 1000000000
#define MOD MM+7
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
template<typename T>
T LCM(T x,T y){
    return x*y/(T)GCD(x,y);
}
int main(){
    int n,ans = 0; cin >> n;
    ans += (n/25);
    n = n%25;
    ans += (n/10);
    n = n%10;
    ans += (n/5);
    n = n%5;
    ans += n;
    cout << ans << endl;
}
  



### 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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <stdio.h>
#include <bitset>
#include <complex>
#include <iomanip>
using namespace std;
#define MM 1000000000
#define MOD MM+7
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
template<typename T>
T LCM(T x,T y){
    return x*y/(T)GCD(x,y);
}
int main(){
    int n,ans = 0; cin >> n;
    ans += (n/25);
    n = n%25;
    ans += (n/10);
    n = n%10;
    ans += (n/5);
    n = n%5;
    ans += n;
    cout << ans << endl;
}
  



```