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 <math.h>

int main(void){
  int N;//県の数
  int M;//市の数
  int P[100000] = {};//市iは県P[i]に属する
  int Y[100000] = {};//市iの誕生年
  int x[100000] = {};//市iは県P[i]においてx[i]番目に誕生した
  int id[100000][12] = {};//id[i]は市[i]の認識番号
  int i, j, k;
  int min;

  scanf("%d %d", &N, &M);
  for(i = 0; i < M; i++){
    scanf("%d %d", &P[i], &Y[i]);
  }

  //これでは計算時間がかかりすぎる
  for(i = 0; i < M; i++){
    for(j = 0; j < M; j++){
      if(P[i] == P[j] && Y[i] >= Y[j]){
        x[i]++;
      }
    }
  }

  for(i = 0; i < M; i++){
    for(j = 0; j < 6; j++){
      id[i][11-j] = x[i]%10 + '0';
      x[i] /= 10;
      id[i][5-j] = P[i]%10 + '0';
      P[i] /= 10;
    }

    for(j = 0; j < 12; j++){
      printf("%c", id[i][j]);
    }
    printf("\n");
  }

  return 0;
}