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

const int N = 10010; // 5000*2+10
int primes[N], cnt;
bool st[N];
int a[N], b[N];

void get_primes(int n) {
    for (int i = 2; i <= n; i++) {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] * i <= n; j++) {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int get(int n, int p) { // n!中p的次数
    int s = 0;
    while (n) n /= p, s += n;
    return s;
}

void mul(int a[], int b, int &len) {
    int t = 0;
    for (int i = 1; i <= len; i++) {
        t += a[i] * b;
        a[i] = t % 10;
        t /= 10;
    }
    while (t) {
        a[++len] = t % 10;
        t /= 10;
    }
    //去前导0
    while (len > 1 && !a[len]) len--;
}

int C(int a, int b, int c[]) {
    int len = 1;
    c[1] = 1;
    for (int i = 0; i < cnt; i++) {
        int p = primes[i];
        int s = get(a, p) - get(b, p) - get(a - b, p);
        while (s--) mul(c, p, len);
    }
    return len;
}

void sub(int a[], int b[], int &len) {
    int t = 0;
    for (int i = 1; i <= len; i++) {
        t = a[i] - b[i] - t;
        a[i] = (t + 10) % 10;
        t < 0 ? t = 1 : t = 0;
    }
    while (len > 1 && !a[len]) len--;
}

int main() {
    //加快读入
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    get_primes(N - 10);

    int n;
    cin >> n;
    int a1 = C(n + n, n, a);
    int b1 = C(n + n, n - 1, b); // bl下面没有用到，原因是两数相减，我们知道a>b,按着a的长度来就行了

    sub(a, b, a1);

    for (int i = a1; i >= 1; i--) printf("%d", a[i]);
    return 0;
}