#define _CRT_SECURE_NO_DepRECATE
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <iostream>
#include <cmath>
#include <iomanip>
#include <string>
#include <algorithm>
#include <bitset>
#include <cstdlib>
#include <cctype>
#include <iterator>
#include <vector>
#include <cstring>
#include <cassert>
#include <map>
#include <queue>
#include <set>
#include <stack>
#define ll long long
#define INF 0x3f3f3f3f
#define ld long double
const ld pi = acos(-1.0L), eps = 1e-8;
int qx[4] = { 0,0,1,-1 }, qy[4] = { 1,-1,0,0 }, qxx[2] = { 1,-1 }, qyy[2] = { 1,-1 };
using namespace std;
struct node
{
    string name;
    int need, end;
}book[20];
struct fate
{
    int last, now, score, day;
}dp[1<<15];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        cin >> n;
        for (int i = 0; i < n; i++)
        {
            cin >> book[i].name >> book[i].end >> book[i].need;
        }
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i < 1 << n; i++)//枚举做作业的每种情况
        {
            dp[i].score = INF;
            for (int f = n - 1; f >= 0; f--)
            {
                int s = 1 << f;//让1代表选择做哪一门课
                if (s & i)//判断该情况是否有做这一门课
                {
                    int last = i - s;//last为没做s这门课的时候
                    int score = max(dp[last].day + book[f].need - book[f].end, 0);//计算做s这门课的时候的分数，分数不能小于0
                    if (score + dp[last].score < dp[i].score)//如果做s这门课得分更少则做这么课
                    {
                        dp[i].score = score + dp[last].score;
                        dp[i].day = dp[last].day + book[f].need;
                        dp[i].last = last;//记录得分最少的上一种情况，记录路径
                        dp[i].now = f;//记录此时选做哪一门课，注意是倒着选的
                    }
                }
            }
        }
        cout << dp[(1 << n) - 1].score << endl;//(1 << n) - 1的情况即为做全部作业的时候，DP的思想。
        int out = (1 << n) - 1;
        stack<string>output;
        while (out)
        {
            output.push(book[dp[out].now].name);//因为是倒着记录的所以应用stack记录然后输出
            out = dp[out].last;
        }
        while (!output.empty())
        {
            cout << output.top() << endl;
            output.pop();
        }
    }
    return 0;
}