There is data on sales of your company. Your task is to write a program which identifies good workers.

The program should read a list of data where each item includes the employee ID i, the amount of sales q and the corresponding unit price p. Then, the program should print IDs of employees whose total sales proceeds (i.e. sum of p × q) is greater than or equal to 1,000,000 in the order of inputting. If there is no such employees, the program should print "NA". You can suppose that n < 4000, and each employee has an unique ID. The unit price p is less than or equal to 1,000,000 and the amount of sales q is less than or equal to 100,000.



Input

The input consists of several datasets. The input ends with a line including a single 0. Each dataset consists of:


n (the number of data in the list)
i p q
i p q
:
:
i p q


Output

For each dataset, print a list of employee IDs or a text "NA"

Example

Input

4
1001 2000 520
1002 1800 450
1003 1600 625
1001 200 1220
2
1001 100 3
1005 1000 100
2
2013 5000 100
2013 5000 100
0


Output

1001
1003
NA
2013
#include <bits/stdc++.h>
using namespace std;
int main() {
  int num;
  int data_set = 0;
  bool do_continue = true;
  std::vector<string> result;
  while (do_continue) {
    cin >> num;
    if ((num == 0) || (data_set > 50)) {
      do_continue = false;
    } else {
      bool na_flg = true;
      data_set++;
      std::map<int, int> e;
      std::vector<int> employee;
      int fst, snd, thd;
      for (int i = 0; i < num; i++) {
        cin >> fst >> snd >> thd;
        employee.push_back(fst);
        if (e.count(employee[i])) {
          e[employee[i]] += snd * thd;
        } else {
          e[employee[i]] = snd * thd;
        }
      }
      for (unsigned int j = 0; j < employee.size(); j++) {
        if (e[employee[j]] >= 1000000) {
          na_flg = false;
          result.push_back(to_string(employee[j]));
        }
      }
      if (na_flg) {
        result.push_back("NA");
      }
    }
  }
  for (unsigned int i = 0; i < result.size() - 1; i++) {
    cout << result[i] << endl;
  }
  return 0;
}
