Problem:
Score: 300 points

Problem Statement

In Republic of Atcoder, there are N prefectures, and a total of M cities that belong to those prefectures.

City i is established in year Y_i and belongs to Prefecture P_i.

You can assume that there are no multiple cities that are established in the same year.

It is decided to allocate a 12-digit ID number to each city.

If City i is the x-th established city among the cities that belong to Prefecture i, the first six digits of the ID number of City i is P_i, and the last six digits of the ID number is x.

Here, if P_i or x (or both) has less than six digits, zeros are added to the left until it has six digits.

Find the ID numbers for all the cities.

Note that there can be a prefecture with no cities.

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq P_i \leq N

1 \leq Y_i \leq 10^9

Y_i are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
P_1 Y_1
:
P_M Y_M

Output

Print the ID numbers for all the cities, in ascending order of indices (City 1, City 2, ...).

Sample Input 1

2 3
1 32
2 63
1 12

Sample Output 1

000001000002
000002000001
000001000001

As City 1 is the second established city among the cities that belong to Prefecture 1, its ID number is 000001000002.

As City 2 is the first established city among the cities that belong to Prefecture 2, its ID number is 000002000001.

As City 3 is the first established city among the cities that belong to Prefecture 1, its ID number is 000001000001.

Sample Input 2

2 3
2 55
2 77
2 99

Sample Output 2

000002000001
000002000002
000002000003

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.text.NumberFormat;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String args[]) {
		Scanner s = new Scanner(System.in);
		  
		  int N;
		  int M;
		  N = s.nextInt();
		  M = s.nextInt();
		  int code[][] = new int[M][4];
		  for(int i=0;i<M;i++) {
			  code[i][0]=i; //原始顺序
			  code[i][1]=s.nextInt();//城市
			  code[i][2]=s.nextInt();//创建时间
			  code[i][3]=0;//排序结果
		  }
		  int order[]= new int[M];
		  int value[]= new int[M];
		  int value_copy[]=new int[M];
		  
		  int l;
		  
		  for(int i=1;i<=N;i++) {
			  l=0;
			  for(int j=0;j<M;j++){
				  if(code[j][3]!=0) {
					  continue;
				  }
				  if(code[j][1]==i) {
					 order[l]=j;
					 value[l]=code[j][2];
					 l++;
				  }
			  }
			  if(l==0) {
				  continue;
			  }
			  value_copy=Arrays.copyOf(value,value.length);			
			  Arrays.sort(value_copy, 0, l);
			  //Arrays.sort(value_copy);
			  for(int x=0;x<M;x++) {
				  if(value[x]==0) {
					  break;
				  }
				  for(int y=0;y<M;y++) {
					  if(value_copy[y]==value[x]){
						  value[x]=y+1;
						  //value[x]=y+l+1-M;
						  break;
					  }
				  }
			  }
			  for(int k=0;k<l;k++) {
				  
				  code[order[k]][3]=value[k];//排序结果
			  }
			  
		  }
		  
		  double resurt;
		  double tt;
	
			NumberFormat nf = NumberFormat.getInstance();
			nf.setGroupingUsed(false);
			nf.setMaximumIntegerDigits(12);	
			nf.setMinimumIntegerDigits(12);
				
		  for(int p=0;p<M;p++) {
			 
			  tt=(int)code[p][1];
			  //System.out.println(tt);
			  resurt = tt*1000000 + (int)code[p][3];
			  //System.out.print(code[p][1]+"\t");
			  //System.out.println(code[p][3]);
			  System.out.println(nf.format(resurt));
		  }
		
			  s.close();
		  
	}
}