1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cmath>
 5 #define LL long long
 6 using namespace std;
 7 const int MAXN = 11;
 8 const int Mod = 9973;
 9 int N;
10 struct mat
11 {
12     int m[MAXN][MAXN];
13 }base;
14 
15 mat muti(mat a, mat b)
16 {
17     mat res;
18     memset(res.m, 0, sizeof(res.m));
19     for(int i = 1; i <= N; i++)
20     for(int j = 1; j <= N; j++){
21             if(a.m[i][j]){
22                 for(int k = 1; k <= N; k++){
23                     res.m[i][k] = (res.m[i][k] + a.m[i][j]*b.m[j][k])%Mod;
24                 }
25             }
26         }
27 
28     return res;
29 }
30 
31 mat qpow(mat a, int n)
32 {
33     mat res;
34     memset(res.m, 0, sizeof(res.m));
35     for(int i = 1; i <= N; i++) res.m[i][i] = 1;
36     while(n){
37         if(n&1) res = muti(res, a);
38         n>>=1;
39         a = muti(a, a);
40     }
41     return res;
42 }
43 
44 
45 int main()
46 {
47     int K, T_case;
48     scanf("%d", &T_case);
49     while(T_case--){
50         memset(base.m, 0, sizeof(base.m));
51         scanf("%d %d", &N, &K);
52         for(int i = 1; i <= N; i++){
53             for(int j = 1; j <= N; j++){
54                 scanf("%d", &base.m[i][j]);
55             }
56         }
57         base = qpow(base, K);
58         int ans = 0;
59         for(int i = 1; i <= N; i++){
60             ans = (ans + base.m[i][i])%Mod;
61         }
62         printf("%d\n", ans);
63     }
64 
65     return 0;
66 }