1 #include <stdio.h>
 2 #include <string.h>
 3 const int maxn = 31;
 4 struct mat
 5 {
 6     unsigned m[maxn][maxn];
 7 }a, E;
 8 unsigned n, k, MOD;
 9
10 mat operator *(mat u, mat v)
11 {
12     mat c;
13     memset(c.m, 0, sizeof c.m);
14     for(int i = 0; i < n; i ++)
15         for(int j = 0; j < n; j ++){
16             for(int r = 0; r < n; r ++)
17                 c.m[i][j] += u.m[i][r] * v.m[r][j];
18             c.m[i][j] %= MOD;
19         }
20     return c;
21 }
22
23 mat operator +(mat u, mat v)
24 {
25     mat c;
26     for(int i = 0; i < n; i ++)
27         for(int j = 0; j < n; j ++)
28             c.m[i][j] = (u.m[i][j] + v.m[i][j]) % MOD;
29     return c;
30 }
31
32 mat cal(mat A, int x)
33 {
34     mat t = E;
35     for(; x; x >>= 1, A = A * A)
36         if(x & 1) t = t * A;
37     return t;
38 }
39
40 mat sum(int x)
41 {
42     if(x == 1)
43         return a;
44     if(x & 1)
45         return cal(a, x) + sum(x - 1);
46     else
47         return sum(x / 2) * (E + cal(a, x / 2));
48     return a;
49 }
50 int main()
51 {
52     while(scanf("%d%d%d", &n, &k, &MOD) != EOF)
53     {
54         for(int i = 0; i < n; i ++)
55         {
56             for(int j = 0; j < n; j ++)
57             {
58                 scanf("%u", &a.m[i][j]);
59                 a.m[i][j] %= MOD;
60                 E.m[i][j] = (i == j);
61             }
62         }
63         mat ans = sum(k);
64         for(int i = 0; i < n; i ++)
65         {
66             for(int j = 0; j < n-1; j ++)
67                 printf("%u ", ans.m[i][j]);
68             printf("%u\n", ans.m[i][n - 1]);
69         }
70     }
71     return 0;
72 }