1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 #define clc(a,b) sizeof(a,b,sizeof(a))
 6 #define LL long long
 7 #include<cmath>
 8 using namespace std;
 9 int p[1010],d[1010];
10 int f[25][4010];//选i个人评审差为j时，评审和最大值。转移方程：f[i][j]=f[i-1][x]+d[k]+p[k];且x+p[k]-d[k]=j;{f[i-1][x]}满足条件的最大值.
11 int path[25][4010];//记录i个人评审差为j,这种情况下的前一个人i-1是谁;
12 int ans[25];
13 int main() {
14 //    freopen("in.txt","r",stdin);
15     int n,m;
16     int grade;
17     int casee=0;
18     while(scanf("%d%d",&n,&m),n||m) {
19         casee++;
20         for(int i=1; i<=n; i++) {
21             scanf("%d%d",&p[i],&d[i]);
22         }
23 //        clc(f,-1);
24 //        clc(path,0);
25         memset(f,-1,sizeof(f));
26         memset(path,0,sizeof(path));
27         grade=m*20;//避免下标为负,所以推广到零以上
28         f[0][grade]=0;//初始化条件
29         for(int i=0; i<=m; i++) {
30             for(int j=0; j<=grade*2; j++) {
31                 if(f[i][j]>=0) {
32                     for(int k=1; k<=n; k++) {
33                         if(f[i][j]+d[k]+p[k]>f[i+1][j+p[k]-d[k]]) {
34                             int t1=i,t2=j;
35                             while(t1>0&&path[t1][t2]!=k) {
36                                 t2-=p[path[t1][t2]]-d[path[t1][t2]];
37                                 t1--;
38                             }//判断k是不是前面已经选过了
39                             if(t1==0) {
40                                 f[i+1][j+p[k]-d[k]]=f[i][j]+d[k]+p[k];
41                                 path[i+1][j+p[k]-d[k]]=k;
42                             }
43                         }
44                     }
45                 }
46             }
47         }
48         int i=grade;
49         int j=0;
50         int k;
51 //        printf("%d\n",f[m][42]);
52         while(f[m][i+j]<0&&f[m][i-j]<0) {
53             j++;
54         }//选出评审分绝对值最小
55         if(f[m][i+j]>f[m][i-j])
56             k=i+j;
57         else
58             k=i-j;
59         printf("Jury #%d\n",casee);
60         printf("Best jury has value %d for prosecution and value %d for defence:\n",(k-grade+f[m][k])/2,(f[m][k]-k+grade)/2);
61         for(i=1; i<=m; i++) {
62             ans[i]=path[m+1-i][k];
63             k-=p[ans[i]]-d[ans[i]];
64         }
65         sort(ans+1,ans+m+1);
66         for(i=1; i<=m; i++) {
67             printf(" %d",ans[i]);
68         }
69         printf("\n\n");
70     }
71 }