Method 1:扩展欧几里得求逆元
-------------------------------------------AC code-------------------------------------------

#include<cstdio>
#include<iostream>

#ifdef WIN32
#   define outl "%I64d"
#else
#   define outl "%lld"
#endif

using namespace std;

typedef long long LL;
const int MOD = (int)1e9+7;
const int N = 1000000;
int T, n;
LL fac[N<<1+5];//fac --> factorial

LL e_gcd(LL a, LL b, LL& x, LL& y){//扩展欧几里得求逆元 
    if(b == 0){
        x = 1, y = 0;
        return a;
    }
    LL ans = e_gcd(b, a%b, x, y);
    LL t = x;
    x = y;
    y = t - (a/b) * y;
    return ans;
}

LL cal_inv(LL a, LL m){
    LL x, y;
    LL gcd = e_gcd(a, m, x, y);
    if(1 % gcd != 0)    return -1;
    x = x * 1 / gcd;
    if(m < 0)   m = -m;
    LL ans = x % m;
    if(ans <= 0)    ans += m;
    return ans;
}

int main(){
    scanf("%d", &T);
    fac[0] = 1;
    for(int i = 1; i <= 2*N; i++)   fac[i] = (fac[i-1] * i) % MOD;
    while(T--){
        scanf("%d", &n);
        LL a = cal_inv(fac[n]*fac[n], MOD);
        printf(outl"\n", fac[2*n] * (a%MOD) % MOD);//a / b % c = a * b1 % c(b1是b在模c意义下的逆元)
    }
    return 0;
}

Method 2:快速幂求逆元
---------------------------------------------AC code---------------------------------------------

#include<cstdio>

#ifdef WIN32
#   define outl "%I64d"
#else
#   define outl "%lld"
#endif

using namespace std;

typedef long long LL;
const int MOD = (int)1e9+7;
const int N = 1000000;
int T, n;
LL fac[N<<1+5];

LL f_pow(LL bs, LL idx, LL mo){
    LL ans = 1;
    bs %= mo;
    while(idx){
        if(idx&1)   ans = ans * bs % mo;
        bs = bs * bs % mo;
        idx >>= 1;
    }
    return ans % mo;
}

int main(){
    scanf("%d", &T);
    fac[0] = 1;
    for(int i = 1; i <= (N<<1); i++)    fac[i] = (fac[i-1]*i) % MOD;
    while(T--){
        scanf("%d", &n);
        printf(outl"\n", fac[n<<1] * f_pow(fac[n]*fac[n], MOD-2, MOD) % MOD);
    }
    return 0;
}