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#include <bits/stdc++.h>
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using namespace std;
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const int mod = 1000007;
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const int N = 410;
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int C[N][N];
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int n, m, k, ans;
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/*
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Sample Input
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2
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2 2 1
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2 3 2
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Sample Output
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Case 1: 0
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Case 2: 2
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*/
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("UVA11806.in", "r", stdin);
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#endif
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// 预处理出组合数结果数组
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for (int i = 1; i < N; i++) {
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C[i][0] = C[i][i] = 1;
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for (int j = 1; j < i; j++)
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C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
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}
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int T, cas = 1;
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int S, s1, s2, s3, s4;
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cin >> T;
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while (T--) {
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ans = 0; // 多组测试数据，每次注意清零
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cin >> n >> m >> k; // n行,m列,k个人
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if (k == 0) { // 一定要注意边界情况，比如0个人
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printf("Case %d: 0\n", cas++);
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continue;
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}
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S = C[n * m][k]; // n*m个格子中找出k个格子站人，就是所有方案数
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s1 = 2 * (C[n * (m - 1)][k] + C[(n - 1) * m][k]) % mod;
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s2 = (C[(n - 2) * m][k] + 4 * C[n * m - n - m + 1][k] + C[n * (m - 2)][k]) % mod;
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s3 = 2 * (C[(n - 2) * (m - 1)][k] + C[(n - 1) * (m - 2)][k]) % mod;
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s4 = C[(n - 2) * (m - 2)][k] % mod;
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ans = (S - s1 + s2 - s3 + s4) % mod; // 容斥原理
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ans = (ans + mod) % mod; // 防止取余后出现负数
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printf("Case %d: %d\n", cas++, ans); // 输出答案
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}
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return 0;
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}
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