#include <set>
#include <map>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define clr(str, x) memset(str, x, sizeof(str))
#define FRER() freopen("in.txt", "r", stdin);
#define FREW() freopen("out.txt", "w", stdout);
#define MAX_INF 0x7fffffff
#define INF 0x3f3f3f3f
#define maxn

typedef long long int ll;
using namespace std;
const int mod = 1000007;
int C[405][405];
int n, m, k, ans = 0;
void init() {
    for (int i = 1; i <= 400; i++) {
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; j++)
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
}
int main() {
    //    FRER()
    //    FREW()
    init();
    int T, kase = 1;
    int s1, s2, s3, s4, s5;
    scanf("%d", &T);
    while (T--) {
        ans = 0;
        scanf("%d%d%d", &n, &m, &k);
        if (k == 0) {
            printf("Case %d: 0\n", kase++);
            continue;
        }

        s1 = C[n * m][k];
        s2 = 2 * (C[n * (m - 1)][k] + C[(n - 1) * m][k]) % mod;
        s3 = (C[(n - 2) * m][k] + 4 * C[n * m - n - m + 1][k] + C[n * (m - 2)][k]) % mod;
        s4 = 2 * (C[(n - 2) * (m - 1)][k] + C[(n - 1) * (m - 2)][k]) % mod;
        s5 = C[(n - 2) * (m - 2)][k] % mod;

        ans = s1 - s2 + s3 - s4 + s5;
        ans = (ans + 10 * mod) % mod; // 防止取余后出现负数
        printf("Case %d: %d\n", kase++, ans);
    }
    return 0;
}