题意：Farmer John owns a farm. He first builds a circle fence. Then, he will choose n points and build some straight fences connecting them. Next, he will feed a cow in each region so that cows cannot play with each other without breaking the fences. In order to feed more cows, he also wants to have as many regions as possible. However, he is busy building fences now, so he needs your help to determine what is the maximum number of cows he can feed if he chooses these n points properly.

思路：

欧拉定理：f = e - v + 2

v = n + C（n，4）（每4个点的连线必有一个交点）

e = n（圆弧） + C（n，2）+ 2*C（n，4）（一个交点是两条边形成的，这个交点将这两条边划分成4条边，数量上+2，因此有多少交点就增加交点数*2条边）

ans = f - 1,圆外的不算

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;
const int mod = 1e9 + 7;

LL power_mod(LL a,LL b)
{
    LL base=a%mod;
    a=a%mod;
    LL res=1LL;
    while(b)
    {
        if(b&1)
            res=base*res%mod;
        base=base*base%mod;
        b=b>>1;
    }
    return res%mod;
}

LL C(LL n, LL m)
{
    if(m > n) return 0;
    LL ans = 1;
    for(int i=1; i<=m; i++)
    {
        LL a = (n + i - m) % mod;
        LL b = i % mod;
        ans = ans * (a * power_mod(b, mod-2) % mod) % mod;
    }
    return ans;
}

LL Lucas(LL n, LL m)
{
    if(m == 0) return 1;
    return C(n % mod, m % mod) * Lucas(n / mod, m / mod) % mod;
}

int main() {
    int T;
    scanf("%d", &T);
    for (int kase = 1; kase <= T; kase++) {
        LL n;
        scanf("%lld", &n);
        printf("Case #%d: %lld\n", kase, (Lucas(n, 2) + Lucas(n, 4) + 1) % mod);
    }
    return 0;
}