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题意：一个宿舍中又n个人，最少k（k >= 3）个人就可以建一个讨论组，问最多可以建多少个不同的讨论组。

解答：C(n,k)+C(n,k+1)+.......+C(n,n)  =  2^n - ( C(n,0) + C(n,1) + C(n,2) + ......+C(n,k-1) )

知识点：费马小定理求逆元inv(a) = qpow( a, mod-2 )。 阶乘逆元：https://blog.csdn.net/Tc_To_Top/article/details/51203160

//C(n, k) + C(n, k + 1) + ....... + C(n, n) = 2 ^ n - (C(n, 0) + C(n, 1) + C(n, 2) + ...... + C(n, k - 1))
#define MAXN 100005
#define mod 1000000007
int t;
int n, k;
ll inv[MAXN];
ll fac[MAXN], coeff[MAXN];
ll QPow(ll x, ll n){ //快速幂计算
    ll ret = 1;
    ll tmp = x % mod;
    while (n){
        if (n & 1) {
            ret = (ret * tmp) % mod;
        }
        tmp = tmp * tmp % mod;
        n >>= 1;
    }
    return ret;
}
void init(){
    fac[0] = 1;
    //fac数组是阶乘
    for (int i = 1; i < MAXN; i++) {
        fac[i] = fac[i - 1] * i % mod;
    }
    
    //   费马小定理求逆元inv(a) = qpow( a, mod-2 )
    //   inv数组存储k阶乘关于模mod的逆元
    inv[MAXN - 1] = QPow(fac[MAXN - 1], mod - 2);
    for (int i = MAXN - 2; i >= 0; i--) {
        //inv[i] = inv[i + 1] * (i + 1) % mod;
        inv[i] = QPow(fac[i], mod - 2);         
    }
}
int main() {
    cin >> t;
    init();
    for(int tn = 1;tn <= t; tn++) {
        cin >> n >> k;
        ll ans = (QPow(2, n) - 1 + mod) % mod;
        coeff[0] = 1;
        for (int i = 1; i < k; i++) {
            //  (a / b) % mod = a *(b关于模mod的逆元)% mod;
            //   C(n,k)=(n-k+1)/k * C(n,k-1)  k从1走到k-1                        
            // C(n,1)=C(n,0)*(n-1+1)/1   C(n,2)=C(n,1)*(n-2+1)/2  C(n,3)=C(n,2)*(n-3+1)/3
            //  令a =coeff[k]=(n-k+1) * C(n,k-1)    b=k
            //  C(n,k) = coeff[k] / k!
            
            coeff[i] = coeff[i - 1] * (n - i + 1) % mod;
            ans = (ans - coeff[i] * inv[i] % mod + mod) % mod;
        }
        printf("Case #%d: %lld\n", tn, ans);
    }
    return 0;
}