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;
struct data {
  int ID;
  long total;
  int order;
  data() {
    ID = -1;
    total = 0;
    order = 100000;
  }
  data(int n) {
    ID = n;
    total = 0;
    order = 100000;
  }
  bool operator<(const data d) const { return order < d.order; }
};
int main() {
  int n;
  while (cin >> n, n) {
    int ID, price, sale;
    data table[4096];
    bool flag = false;
    int order_count = 0;
    getchar();
    for (int i = 0; i < sizeof(table) / sizeof(table[0]); i++) {
      table[i] = data(i);
    }
    for (int i = 0; i < n; i++) {
      char in[128];
      char ID_s[32], price_s[32], sale_s[32];
      int j, k;
      fgets(in, 120, stdin);
      for (j = 0, k = 0; in[j + k] != ' ' && in[j + k] != ';'; k++)
        ID_s[k] = in[j + k];
      ID_s[k] = '\0';
      j = k + 1;
      for (k = 0; in[j + k] != ' ' && in[j + k] != ';'; k++)
        price_s[k] = in[j + k];
      price_s[k] = '\0';
      j += k + 1;
      for (k = 0; in[j + k] != ' ' && in[j + k] != ';' && in[j + k] &&
                  in[j + k] != '\n';
           k++)
        sale_s[k] = in[j + k];
      sale_s[k] = '\0';
      ID = atoi(ID_s);
      price = atoi(price_s);
      sale = atoi(sale_s);
      if (table[ID].order == 100000) {
        table[ID].order = order_count;
        order_count++;
      }
      table[ID].total += price * sale;
    }
    sort(table, table + 4096);
    for (int i = 0; i < order_count; i++) {
      if (table[i].total >= 1000000) {
        cout << table[i].ID << endl;
        flag = true;
      }
    }
    if (!flag) cout << "NA" << endl;
  }
  return 0;
}
