#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

LL C(int n, int m) {
    LL a = 1, b = 1;
    for (int i = 0; i < m; i++) {
        a *= (n - i);
        b *= (m - i);
    }
    return a / b;
}
int main() {
    int n;
    cin >> n; // n过大会溢出，需要结合高精度
    printf("%lld\n", C(2 * n, n) / (n + 1));
    return 0;
}