#pragma region include
#include <iostream>
#include <iomanip>
#include <stdio.h>

#include <sstream>
#include <algorithm>
#include <cmath>
#include <complex>

#include <string>
#include <cstring>
#include <vector>
#include <tuple>
#include <bitset>

#include <queue>
#include <complex>
#include <set>
#include <map>
#include <stack>
#include <list>

#include <fstream>
#include <random>
//#include <time.h>
#include <ctime>
#pragma endregion //#include
/////////
#define REP(i, x, n) for(int i = x; i < n; ++i)
#define rep(i,n) REP(i,0,n)
/////////
#pragma region typedef
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
#pragma endregion //typedef
////定数
const int INF = (int)1e9;
const LL MOD = (LL)1e9+7;
const LL LINF = (LL)1e18;
const double PI = acos(-1.0);
const double EPS = 1e-9;
/////////
using namespace::std;
/////////
#pragma region Math
#pragma region
template<class T>
inline T gcd(T a, T b){return b ? gcd(b, a % b) : a;}
#pragma endregion // 最大公約数 gcd
#pragma region
template<class T>
inline T lcm(T a, T b){return a / gcd(a, b) * b;}
#pragma endregion // 最小公倍数 lcm
#pragma region
LL powMod(LL num,LL n,LL mod=(LL)MOD){//(num**n)%mod
	num %= mod;//
	if( n == 0 ){
		return (LL)1;
	}
	LL mul = num;
	LL ans = (LL)1;
	while(n){
		if( n&1 ){
			ans = (ans*mul)%mod;
		}
		mul = (mul*mul)%mod;
		n >>= 1;
	}
	return ans;
}
LL mod_inverse(LL num,LL mod=MOD){
	return powMod(num,MOD-2,MOD);
}
#pragma endregion //繰り返し二乗法 powMod

void solve(){
	LL N;
	cin >> N;
	LL inv = mod_inverse(2);
	LL invN = powMod(inv,N);
	LL temp = 2*N;
	LL ans=1;
	for(int i=1;i<=temp;++i){
		ans = (ans * i) % MOD;
	}
	ans = (ans * invN ) % MOD;

	cout << ans << endl;
}

#pragma region main
signed main(void){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	std::cout << std::fixed;//小数を10進数表示
	cout << setprecision(16);//小数点以下の桁数を指定//coutとcerrで別	
	
	solve();
}
#pragma endregion //main()