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#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;
