1 //#define LOCAL
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 using namespace std;
 7
 8 const int maxn = 5000 + 10;
 9 const int INF = 1000000000;
10 char p[maxn], q[maxn];
11 int sp[26], ep[26], sq[26], eq[26];    //记录两个串每种颜色的始末
12 int d[2][maxn], c[2][maxn];    //c[i][j]表示d[i][j]时还有多少种颜色已经开始但尚未结束
13
14 int main(void)
15 {
16     #ifdef LOCAL
17         freopen("1625in.txt", "r", stdin);
18     #endif
19
20     int n, m, T;
21     scanf("%d", &T);
22     while(T--)
23     {
24         scanf("%s", p+1);
25         scanf("%s", q+1);
26         n = strlen(p+1);
27         m = strlen(q+1);
28         for(int i = 1; i <= n; ++i)    p[i] -= 'A';
29         for(int i = 1; i <= m; ++i)    q[i] -= 'A';
30
31         for(int i = 0; i < 26; ++i)    { sp[i] = sq[i] = INF; ep[i] = eq[i] = 0; }
32         for(int i = 1; i <= n; ++i)
33         {
34             sp[p[i]] = min(sp[p[i]], i);
35             ep[p[i]] = i;
36         }
37         for(int i = 1; i <= m; ++i)
38         {
39             sq[q[i]] = min(sq[q[i]], i);
40             eq[q[i]] = i;
41         }
42
43         int t = 0;
44         memset(d, 0, sizeof(d));
45         memset(c, 0, sizeof(c));
46         for(int i = 0; i <= n; ++i)
47         {
48             for(int j = 0; j <= m; ++j)
49             {
50                 if(!i && !j)    continue;
51
52                 int v1, v2;
53                 v1 = v2 = INF;
54                 if(i)    v1 = d[t^1][j] + c[t^1][j];
55                 if(j)    v2 = d[t][j-1] + c[t][j-1];
56                 d[t][j] = min(v1, v2);
57                 //计算c
58                 if(i)
59                 {
60                     c[t][j] = c[t^1][j];
61                     if(sp[p[i]] == i && sq[p[i]] > j)    c[t][j]++;
62                     if(ep[p[i]] == i && eq[p[i]] <= j)    c[t][j]--;
63                 }
64                 else if(j)
65                 {
66                     c[t][j] = c[t][j-1];
67                     if(sq[q[j]] == j && sp[q[j]] > i)    c[t][j]++;
68                     if(eq[q[j]] == j && ep[q[j]] <= i)    c[t][j]--;
69                 }
70             }
71             t ^= 1;
72         }
73         printf("%d\n", d[t^1][m]);
74     }
75
76     return 0;
77 }