Problem:
Sale Result

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"

Sample 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 for the Sample Input

1001
1003
NA
2013

Solution in C (judge verdict: Accepted)

#include<stdio.h>

int main(){
	int kazu;					/*f[^yÑAüÍÌIíè*/
	long long data[4000][4];	/*{ÐõÔ,ÌP¿,ãÊ,èã°Ìg[^}*/
	int kaunto[4000][2];		/*{ÐõÔ,üÍ³ê½Ô}*/
	int count;					/*1[vàÅAY·éÐõª¢½©Ç¤©B(0¨³ 1¨L) */
	int i,j;					/*[vpÏ*/
	int number=0;				/*ÐõÔüÍpÏ*/
	
	for(i=0;i<4000;i++){		/*ÐõÔÌú»*/
		data[i][0]=i+1;
		kaunto[i][0]=i+1;
	}
	
	scanf("%d",&kazu);
	while(kazu!=0){
		count=0;
		for(i=0;i<4000;i++){		/*ú»*/
			for(j=1;j<4;j++){
				data[i][j]=0;
			}
		}
		for(i=0;i<4000;i++){
			kaunto[i][1]=0;
		}
		
		for(i=1;i<=kazu;i++){
			scanf("%d",&number);
			scanf("%lld %lld",&data[number-1][1],&data[number-1][2]);	/*ÐõÔÉÎ·ézñÉf[^ðüêÄ¢­B*/
			data[number-1][3]+=data[number-1][1]*data[number-1][2];
			if(kaunto[number-1][1]==0){									/*Ü¾üÍ³êÄ±ÆªÈ©Á½çA*/
				kaunto[number-1][1]=i;									/*üÍÌÔðüêéB*/
			}
		}
		
		for(i=1;i<=4000;i++){	/*üÍÔÌ´ç¢*/
			for(j=0;j<4000;j++){	/*ÐõÔÌ´ç¢*/
				if(kaunto[j][1]==i && data[j][3]>=1000000){	/*(üÍÌÔªiÌàÌª Á½çA(­È¢©ç´ç¢) && g[^ª1,000,000Èã¾Á½çA)*/
					printf("%d\n",kaunto[j][0]);
					count=1;		/*¡ñÌüÍÅY·éÐõª¢½B*/
				}
			}
		}
		if(count==0){				/*¡ñÌüÍÅY·éÐõª¢È©Á½çA*/
			printf("NA\n");
		}
		
		scanf("%d",&kazu);
	}
	return 0;
}