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: Wrong Answer)

#include <stdio.h>
int main(void)
{
  int n, m, i, j, k, l, o, tmp, top;
  scanf("%d %d", &n, &m);

  int p[m], y[m];

  for(i=0;i<m;i++){
    scanf("%d %d",&p[i], &y[i]);
  }

  for(j=0;j<m-1;j++){
    for(k=j+1;k<m;k++){
      if(y[j]<y[k]){
        tmp = y[j];
        y[j] = y[k];
        y[k] = y[j];

        top = p[j];
        p[j] = p[k];
        p[k] = top;
      }
    }
  }

  for(o=0;o<m;o++){
    y[0] = 1;
    if(p[o]==p[o+1]){
      y[o+1] = y[o]+1;
    }
    else{
      y[o+1] = 0;
    }

  }

  for(l=0;l<m;l++){
    printf("%06d%06d\n", p[l], y[l]);
  }

  return 0;

}