#include <bits/stdc++.h>
using namespace std;
const int N = 4010;
const int P = 2147483648;
//f[i][j]:考虑前i个数字,总和是j的方案数
unsigned f[N][N];
int n;
int main()
{
cin >> n;
f[0][0] = 1;
for(int i = 1; i <= n; i++)
{
for(int j = 0; j <= n; j++)
{
//不选
f[i][j] = f[i-1][j];
//选
if(j >= i)
f[i][j] = (f[i][j] + f[i][j-i]) % P;
}
}
cout << (f[n][n] - 1) % P << endl;
return 0;
}