//完全背包  求方案数目

//f[i][j]   只从前i个物品中选，且总体积恰好为j的方案的集合

//f[i][j]=f[i-1][j]+f[i-1][j-v*1]+f[i-1][j-v*2]+...f[i-1][j-v*s]

//f[i][j-v]=        f[i-1][j-v*1]+f[i-1][j-v*2]+...f[i-1][j-v*s]

//所以  f[i][j]=f[i-1][j]+f[i][j-v]

#include <iostream>

using namespace std;

const int N = ;

int n;

int v[] = {, , , };

int f[N];

int main() {

    cin >> n;

    f[] = ;

    for (int i = ; i < ; i ++ )

        for (int j = v[i]; j <= n; j ++ )

            f[j] =f[j] + f[j - v[i]];

    cout << f[n] << endl;

    return ;

}