#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
long long powm(long long a, long long b, long long m = 1000000007) {
  long long ans = 1;
  while (b > 0) {
    if ((b & 1) == 1) ans = (ans * a) % m;
    b >>= 1;
    a = (a * a) % m;
  }
  return ans % m;
}
long long inv_mod(long long a, long long m = 1000000007) {
  return powm(a, m - 2, m);
}
long long ad(long long a, long long b, long long m = 1000000007) {
  a = a + b;
  if (a >= m) a -= m;
  return a;
}
long long sb(long long a, long long b, long long m = 1000000007) {
  a = a - b;
  if (a < 0) a += m;
  return a;
}
long long mul(long long a, long long b, long long m = 1000000007) {
  return (a * b) % m;
}
long long mdiv(long long a, long long b, long long m = 1000000007) {
  return mul(a, inv_mod(b));
}
long long n;
const long long N = 2e6 + 2;
std::vector<long long> first(N);
std::vector<long long> inv(N);
long long ncr(long long i, long long r) {
  if (r > i) return 0;
  long long ret = first[i];
  ret = mul(ret, inv[r]);
  ret = mul(ret, inv[i - r]);
  return ret;
}
void solve() {
  cin >> n;
  first[0] = 1;
  inv[0] = 1;
  std::vector<long long> a(2 * n + 1);
  for (long long i = 1; i < 2 * n + 1; ++i) {
    first[i] = mul(first[i - 1], i);
    inv[i] = inv_mod(first[i]);
  }
  long long ans = 3;
  a[1] = 2;
  for (long long i = 2; i < 2 * n + 1; ++i) {
    long long x = 2 * ncr(i - 1, n);
    a[i] = ad(x, mul(sb(a[i - 1], x), 2));
    ans = ad(ans, a[i]);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long T = 1;
  while (T--) {
    solve();
  }
}
