#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int fix = 400;
int dp[30][1000]; // dp[i][j]表示选了i个人，差值为j的时候，的最大和值
int path[30][1000];
int D[200 + 20];
int P[200 + 20];
int sum[200 + 20];
int dis[200 + 20];
set<int>ans;
int n, m;
void init() {memset(dp, -1, sizeof dp);memset(path, -1, sizeof path);ans.clear();
}
void work() {init();for (int i = 1; i <= n; ++i) {cin >> D[i] >> P[i];dis[i] = D[i] - P[i];sum[i] = D[i] + P[i];}dp[0][fix] = 0; //因为fix了，所以0的时候，其实就是fix的时候for (int i = 1; i <= m; ++i) {for (int k = 0; k <= 2 * fix; ++k) {if (dp[i - 1][k] >= 0) { //如果这个状态可行，才能再选举一个人进来for (int j = 1; j <= n; ++j) {if (dp[i - 1][k] + sum[j] > dp[i][k + dis[j]]) { //加了这个人后，最优解变大，然后差值就是变成k + dis[j]的//还需要检查j是否在dp[i - 1][k]这个路径上了。int t = path[i - 1][k];int pre = i - 1;int ff = k;bool flag = true;while (t != -1) {if (t == j) {flag = false;break;}ff -= dis[t]; //减去当前这个点的值t = path[--pre][ff];}if (flag == true) {dp[i][k + dis[j]] = dp[i - 1][k] + sum[j];path[i][k + dis[j]] = j;}}}}}}int i = fix;int j = 0;while (dp[m][i + j] < 0 && dp[m][i - j] < 0) { //排除没用的状态++j;}int KK; //最小的差值 对应的那个dp，可以对应输出最大的和值if (dp[m][i + j] > dp[m][i - j]) {KK = i + j;} else KK = i - j;int t = path[m][KK];int pre = m;int ff = KK;while (t != -1) {ans.insert(t);ff -= dis[t];t = path[--pre][ff];}int df = 0, pf = 0;for (set<int> :: iterator it = ans.begin(); it != ans.end(); ++it) {df += D[*it];pf += P[*it];}static int gg = 0;printf("Jury #%d\n", ++gg);printf("Best jury has value %d for prosecution and value %d for defence:\n", df, pf);for (set<int> :: iterator it = ans.begin(); it != ans.end(); ++it) {cout << " " << *it;}cout << endl << endl;
}int main() {
#ifdef localfreopen("data.txt","r",stdin);
#endifIOS;while (cin >> n >> m && (n + m)) {work();}return 0;
}