1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 const int Maxn = 1005, INF = 1e9 + 7;
 8 int N, M, T[Maxn], B[Maxn], Dp[Maxn][1 << 8][20];
 9 
10 int main()
11 {
12     scanf("%d", &M);
13     while (M--)
14     {
15         scanf("%d", &N);
16         memset(Dp, 0x7f, sizeof Dp);
17         Dp[1][0][7] = 0;
18         for (int i = 1; i <= N; ++i) scanf("%d %d", T + i, B + i);
19         for (int i = 1; i <= N; ++i)
20             for (int j = 0; j < (1 << 8); ++j)
21                 for (int k = -8; k < 8; ++k)
22                 {
23                     if (Dp[i][j][k + 8] >= INF)continue;
24                     if (j & 1) Dp[i + 1][j >> 1][k + 7] = min(Dp[i + 1][j >> 1][k + 7], Dp[i][j][k + 8]);
25                     else 
26                     {
27                         int Lim = INF;
28                         for (int h = 0; h < 8; ++h)
29                         {
30                             if ((j >> h) & 1)continue;
31                             if (i + h > Lim)break;
32                             Lim = min(Lim,  i + h + B[i + h]);
33                             Dp[i][j | (1 << h)][h + 8] = min(Dp[i][j | (1 << h)][h + 8], Dp[i][j][k + 8] + (i + k ? T[i + k] ^ T[i + h] : 0));
34                         }
35                     }
36                 }
37         int Ans(INF);
38         for (int k = 0; k < 9; ++k) Ans = min(Ans, Dp[N + 1][0][k]);
39         printf("%d\n", Ans);
40     }
41 }