1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef __int64 LL;
15 typedef pair <int, int> P;
16 const int N = 1e6+ 5;
17 LL mod = 1e9 + 7;
18 LL f[N];
19
20 LL Pow(LL a , LL n , LL mod) {
21     LL res = 1;
22     while(n) {
23         if(n & 1)
24             res = res * a % mod;
25         a = a * a % mod;
26         n >>= 1;
27     }
28     return res;
29 }
30
31 LL Cnm(LL n, LL m, LL mod) {
32     if(n < N) { //数据范围小的 逆元就好了
33         return f[n]*Pow(f[m]*f[n-m]%mod, mod - 2, mod) % mod;
34     }
35     LL res = Pow(f[m]%mod, mod - 2, mod)%mod;
36     for(LL i = 0; i < m; ++i) {
37         res = res * (n-i) % mod;
38     }
39     return res%mod;
40 }
41
42 void init() { //阶乘预处理
43     f[0] = 1;
44     for(LL i = 1; i < N; ++i)
45         f[i] = f[i - 1] * i % mod;
46 }
47
48 int main()
49 {
50     init();
51     int t;
52     LL n, m, k;
53     scanf("%d", &t);
54     for(int ca = 1; ca <= t; ++ca) {
55         scanf("%lld %lld %lld", &n, &m, &k);
56         printf("Case #%d: ", ca);
57         if(k == 1 && n == 1) {
58             printf("%lld\n", m);
59             continue;
60         }
61         else if(k == 1) {
62             printf("0\n");
63             continue;
64         }
65         LL ans = 0, temp = k;
66         for(int i = 1; k >= 2; --k, i = -i) {
67             ans = (i * Cnm(temp, k, mod) % mod *k % mod* Pow(k- 1, n - 1, mod) % mod + ans) % mod;
68         }
69         ans = ans * Cnm(m, temp, mod) % mod;
70         printf("%lld\n", (ans + mod) % mod);
71     }
72     return 0;
73 }