1     //#pragma comment(linker, "/STACK:102400000, 102400000")
 2     #include <algorithm>
 3     #include <iostream>
 4     #include <cstdlib>
 5     #include <cstring>
 6     #include <cstdio>
 7     #include <vector>
 8     #include <cmath>
 9     #include <ctime>
10     #include <list>
11     #include <set>
12     #include <map>
13     using namespace std;
14     typedef long long LL;
15     typedef pair <int, int> P;
16     const int N = 4e3 + 5;
17     LL dp[N][5]; //dp[i][k]表示使用i体积有k个木板在露在外面的最大价值
18     int w[N];
19     LL v[N];
20
21     int main()
22     {
23         int t, n, g;
24         scanf("%d", &t);
25         for(int ca = 1; ca <= t; ++ca) {
26             scanf("%d %d", &n, &g);
27             LL ans = 0;
28             for(int i = 1; i <= n; ++i) {
29                 scanf("%d %lld", w + i, v + i);
30                 w[i] <<= 1;
31                 ans = max(ans, v[i]);
32             }
33             g <<= 1;
34             memset(dp, 0, sizeof(dp));
35             for(int i = 1; i <= n; ++i) {
36                 for(int j = g; j >= w[i]/2; --j) { //一维背包类似，说明下面只能取一次
37                     for(int k = 0; k <= 2; ++k) {
38                         if(w[i] <= j) //全放在里面
39                             dp[j][k] = max(dp[j - w[i]][k] + v[i], dp[j][k]);
40                         if(k >= 1) //有一半在外面
41                             dp[j][k] = max(dp[j - w[i]/2][k - 1] + v[i], dp[j][k]);
42                     }
43                 }
44             }
45             for(int i = 0; i <= 2; ++i) {
46                 ans = max(ans, dp[g][i]);
47             }
48             printf("Case #%d: %lld\n", ca, ans);
49         }
50         return 0;
51     }