1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 #include<vector>
 5 using namespace std;
 6
 7 const int mod = 1000003;
 8
 9 int f[51][65536];
10 int t, n, len, k;
11 int ans;
12 vector<int> G[27];       //动态数组
13 char c[16][52];
14
15 void dp() {
16     memset(f, 0, sizeof(f));
17     int num = 0;
18     for (int i = 0; i < 27; i++)  //预先处理第一位
19         G[i].clear();
20     for (int i = 1; i <= n; i++) {
21         if (c[i][0] == '?')
22             G[26].push_back(i);
23         else G[(int)c[i][0] - 97].push_back(i);
24     }
25     int any = 0;
26     for (int i = 0; i < G[26].size(); i++)
27         any = any + (1 << (G[26][i] - 1));   //any表示下一位为？的字符串的匹配状态
28     for (int i = 0; i < 26; i++) {
29         int lsany = any;   //lsany表示分别下一位为a - z时的匹配状体
30         for (int j = 0; j < G[i].size(); j++)
31             lsany = lsany + (1 << (G[i][j] - 1));
32         if (lsany != 0)
33             f[1][lsany] = f[1][lsany] % mod + 1;   //动态转移
34         if (lsany > num) num = lsany;
35     }
36     for (int i = 1; i < len; i++)
37         for (int j = 1; j <= num; j++)
38             if (f[i][j] != 0)  {
39                 for (int k = 0; k < 27; k++)
40                     G[k].clear();
41                 for (int k = 0; k < n; k++)
42                     if ((1 & j >> k) == 1)
43                         if (c[k+1][i] == '?')
44                             G[26].push_back(k+1);
45                         else G[(int) c[k+1][i] - 97].push_back(k+1);
46                 int any = 0;
47                 for (int k = 0; k < G[26].size(); k++)
48                     any = any + (1 << (G[26][k] - 1));
49                 for (int k = 0; k < 26; k++) {
50                     int lsany = any;
51                     for (int w = 0; w < G[k].size(); w++)
52                         lsany = lsany + (1 << (G[k][w] - 1));
53                     if (lsany != 0)
54                         f[i+1][lsany] =(f[i+1][lsany]  + f[i][j])  % mod;
55                 }
56             }
57 }
58
59
60 void print(int now, int where, int w){   //输出
61     if (now == k) {ans =(ans + f[len][w]) % mod; return;}
62     if (where > n) return;
63     print(now+1, where+1, w + (1 << (where - 1)));
64     print(now, where+1, w);
65 }
66
67 int main() {
68     ios::sync_with_stdio;
69     cin >> t;
70     for (int o = 0; o < t; o++) {
71         cin >> n >> k;
72         for (int i = 1; i <= n; i++)
73             cin >> c[i];
74         len = 0;
75         while ((int)c[1][len] > 96 || c[1][len] == '?')
76             len++;
77         dp();
78         ans = 0;
79         print(0, 1, 0);
80         cout << ans << "\n";
81     }
82     return 0;
83 }