一道智障题。。。好吧是我做的很智障B. Qualifying Contest

你看看有些人。就把那些可以用高档技巧牛逼语言的题，做的粗暴的丑不拉几。
好吧就是我。
虽然晚饭跑去吃了水煮鱼已经不难过了。
可看到这个东西还是会多多少少郁闷一下的。

这个题有看到别人用vector或者map做的很好看。
嗯我的弱智struct，，，还是贴上来吧2333

啊郁闷的原因还和上道题一样。因为自己做的很简单啦比赛的时候却没写orz可能是有毒吧。
(⊙v⊙)嗯以及还是要好好学习vector的用法。

题目：

B. Qualifying Contest

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

Very soon Berland will hold a School Team Programming Olympiad. From each of the m Berland regions a team of two people is invited to participate in the olympiad. The qualifying contest to form teams was held and it was attended by n Berland students. There were at least two schoolboys participating from each of the m regions of Berland. The result of each of the participants of the qualifying competition is an integer score from 0 to 800 inclusive.

The team of each region is formed from two such members of the qualifying competition of the region, that none of them can be replaced by a schoolboy of the same region, not included in the team and who received a greater number of points. There may be a situation where a team of some region can not be formed uniquely, that is, there is more than one school team that meets the properties described above. In this case, the region needs to undertake an additional contest. The two teams in the region are considered to be different if there is at least one schoolboy who is included in one team and is not included in the other team. It is guaranteed that for each region at least two its representatives participated in the qualifying contest.

Your task is, given the results of the qualifying competition, to identify the team from each region, or to announce that in this region its formation requires additional contests.

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 10 000, n ≥ 2m) — the number of participants of the qualifying contest and the number of regions in Berland.

Next n lines contain the description of the participants of the qualifying contest in the following format: Surname (a string of length from 1 to 10 characters and consisting of large and small English letters), region number (integer from 1 to m) and the number of points scored by the participant (integer from 0 to 800, inclusive).

It is guaranteed that all surnames of all the participants are distinct and at least two people participated from each of the m regions. The surnames that only differ in letter cases, should be considered distinct.

Output

Print m lines. On the i-th line print the team of the i-th region — the surnames of the two team members in an arbitrary order, or a single character “?” (without the quotes) if you need to spend further qualifying contests in the region.

Examples

Input
5 2
Ivanov 1 763
Andreev 2 800
Petrov 1 595
Sidorov 1 790
Semenov 2 503

Output
Sidorov Ivanov
Andreev Semenov

Input
5 2
Ivanov 1 800
Andreev 2 763
Petrov 1 800
Sidorov 1 800
Semenov 2 503

Output
?
Andreev Semenov

Note

In the first sample region teams are uniquely determined.

In the second sample the team from region 2 is uniquely determined and the team from region 1 can have three teams: “Petrov”-“Sidorov”, “Ivanov”-“Sidorov”, “Ivanov” -“Petrov”, so it is impossible to determine a team uniquely.

代码：

#include"iostream"
#include"string"
#include"string.h"
#include"algorithm"
#include"vector"
#include"queue"
using namespace std;

struct memsage
{
    string name;
    int city=0;
    int grade;
}person[100010];

bool cmp(memsage a,memsage b)
{
    if (a.city < b.city)
        return true;
    else if (a.city == b.city&&a.grade > b.grade)
        return true;
    else
        return false;

}

int main()
{
    int n, m;
    while (cin >> n >> m)
    {
        int num_ecity[10010];
        memset(num_ecity, 0, sizeof(num_ecity));
        for (int i = 0; i < n; i++)
        {
            cin >> person[i].name >> person[i].city >> person[i].grade;
            num_ecity[person[i].city]++;
        }

        sort(person, person + n, cmp);

        int pos = 1;

        for (int i = 0; i < n; i++)
        {
            if ((person[i + 1].grade != person[i + 2].grade) || (person[i + 1].city != person[i + 2].city))
                cout << person[i].name << " " << person[i + 1].name << endl;
            else
                cout << "?" << endl;
            i = i + num_ecity[pos]-1;
            pos++;
        }
    }

    return 0;
}

这个要贴上来的原因主要是struct内部的比较。
每次写这个都要参考一下之前写过的一个程序。
这样当然不太好啦。
所以希望能够记住吧！嘻嘻嘻~

以上。