A registration card number of PAT consists of 4 parts:

• the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic;
• the 2nd - 4th digits are the test site number, ranged from 101 to 999;
• the 5th - 10th digits give the test date, in the form of yymmdd;
• finally the 11th - 13th digits are the testee's number, ranged from 000 to 999.

Now given a set of registration card numbers and the scores of the card owners, you are supposed to output the various statistics according to the given queries.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (≤10​4​​) and M (≤100), the numbers of cards and the queries, respectively.

Then N lines follow, each gives a card number and the owner's score (integer in [0,100]), separated by a space.

After the info of testees, there are M lines, each gives a query in the format Type Term, where

• Type being 1 means to output all the testees on a given level, in non-increasing order of their scores. The corresponding Term will be the letter which specifies the level;
• Type being 2 means to output the total number of testees together with their total scores in a given site. The corresponding Term will then be the site number;
• Type being 3 means to output the total number of testees of every site for a given test date. The corresponding Term will then be the date, given in the same format as in the registration card.

Output Specification:

For each query, first print in a line Case #: input, where # is the index of the query case, starting from 1; and input is a copy of the corresponding input query. Then output as requested:

• for a type 1 query, the output format is the same as in input, that is, CardNumber Score. If there is a tie of the scores, output in increasing alphabetical order of their card numbers (uniqueness of the card numbers is guaranteed);
• for a type 2 query, output in the format Nt Ns where Nt is the total number of testees and Ns is their total score;
• for a type 3 query, output in the format Site Nt where Site is the site number and Nt is the total number of testees at Site. The output must be in non-increasing order of Nt's, or in increasing order of site numbers if there is a tie of Nt.

If the result of a query is empty, simply print NA.

Sample Input:

8 4

B123180908127 99

B102180908003 86

A112180318002 98

T107150310127 62

A107180908108 100

T123180908010 78

B112160918035 88

A107180908021 98

1 A

2 107

3 180908

2 999

Sample Output:

Case 1: 1 A

A107180908108 100

A107180908021 98

A112180318002 98

Case 2: 2 107

3 260

Case 3: 3 180908

107 2

123 2

102 1

Case 4: 2 999

NA

#include <stdio.h>

#include <string>

#include <iostream>

#include <algorithm>

#include <vector>

#include <map>

#include <unordered_map>

using namespace std;

struct stu {

    string s;

    int score;

};

vector<stu> v;

int n, m;

string s;

int score;

bool cmp1(stu s1, stu s2) {

    return s1.score == s2.score ? s1.s < s2.s : s1.score>s2.score;

}

int main() {

    cin >> n >> m;

    for (int i = ; i < n; i++) {

        cin >> s >> score;

        getchar();

        stu s1;

        s1.s = s;

        s1.score = score;

        v.push_back(s1);

    }

    for (int i = ; i <= m; i++) {

        vector<stu> ans;

        string query;

        int num;

        cin >> num >> query;

        printf("Case %d: %d %s\n", i, num, query.c_str());

        int flag = ;

        if (num == ) {

            for (int j = ; j < n; j++) {

                if (v[j].s[] == query[]) {

                    ans.push_back(v[j]);

                    flag = ;

                }

            }

        }

        else if (num == ) {

            int total = , count = ;

            for (int j = ; j < n; j++) {

                if (v[j].s.substr(, ) == query) {

                    total += v[j].score;

                    count++;

                    flag = ;

                }

            }

            if(flag==)printf("%d %d\n", count, total);

        }

        else if (num == ) {

            unordered_map<string, int> mp3;

            for (int j = ; j < n; j++) {

                if (v[j].s.substr(, ) == query) {

                    mp3[v[j].s.substr(, )]++;

                    flag = ;

                }

            }

            if (flag == ) {

                for (auto it:mp3) {

                    ans.push_back({ it.first,it.second });

                }

            }

        }

        sort(ans.begin(), ans.end(), cmp1);

        for (int j = ; j < ans.size(); j++) {

            printf("%s %d\n", ans[j].s.c_str(), ans[j].score);

        }

        if (flag == )printf("NA\n");

    }

    system("pause");

}