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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct hoge{
  int bangou;
  int ken;
  int tosi;
} hoge;

int main(){

  int i, j, n, m;
  hoge aaa[100010];
  char ans[100010][12];
  scanf("%d %d", &n, &m);
  for(i=0;i<m;i++){
    scanf("%d %d", &aaa[i].ken, &aaa[i].tosi);
    aaa[i].bangou = i;
    for(j=0;j<12;j++){
      ans[i][j] = '0';
    }
  }
  for(i=0;i<m;i++){
    for(j=i+1;j<m;j++){
      if(aaa[i].tosi > aaa[j].tosi){
        aaa[100005] = aaa[i];
        aaa[i] = aaa[j];
        aaa[j] = aaa[100005];
      }
    }
  }

  int count[100010]={0};
  int keta, t;

  for(i=0;i<m;i++){
    count[aaa[i].ken]++;
    keta = 5;
    t = aaa[i].ken;
    while(t>0){
      ans[aaa[i].bangou][keta] = t%10+'0';
      t /= 10;
      keta--;
    }
    keta=11;

    t = count[aaa[i].ken];

    while(t>0){
      ans[aaa[i].bangou][keta] = t%10+'0';
      t /= 10;
      keta--;
    }
  }
  int gomi[1000010];
  for(i=0;i<m;i++){
    for(j=0;j<12;j++){
      printf("%c",ans[i][j]);
    }
    putchar('\n');
  }
  return 0;
}