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

typedef struct t_city_node {
  int state;
  int year;
  int seq_id;
} t_city;

void merge(int N, t_city *city, int *idx, int left, int mid, int right) {
  int tmp[100000];
  int i, j, k;

  for(i = left; i <= mid; i++) {
    tmp[i] = idx[i];
  }
  for(i = mid+1, j = right; i <= right; i++, j--){
    tmp[i] = idx[j];
  }
  i = left;
  j = right;
  for(k = left; k <= right; k++) {
    if(city[idx[i]].year == city[idx[j]].year) {
      idx[k] = tmp[i];
    } else if(city[idx[i]].year > city[idx[j]].year) {
      idx[k] = tmp[j--];
    } else {
      idx[k] = tmp[i++];
    }
  }
}

void my_msort(int N, t_city *city, int *idx, int left, int right) {
  if(left >= right) return;
  int mid = (left + right) / 2;
  my_msort(N, city, idx, left, mid);
  my_msort(N, city, idx, mid+1, right);
  merge(N, city, idx, left , mid, right);
}
int main(void) {
  int N, M;
  scanf("%d %d", &N, &M);

  t_city city_list[100000];
  int idx[100000];
  for(int i = 0; i < M; i++) {
    scanf("%d %d", &city_list[i].state, &city_list[i].year);
    idx[i] = i;
  }

  // sort by year
  my_msort(M, city_list, idx, 0, M-1);

  // set seq_id
  int city_id[100000];
  memset(city_id,  0, sizeof(int)*100000);
  for(int i = 0; i < M; i++) {
    int state = city_list[idx[i]].state;
    city_list[idx[i]].seq_id = ++city_id[state];
  }
    
  // final output
  for(int i = 0; i < M; i++) {
    printf("%06d%06d\n",
	   city_list[i].state,
	   city_list[i].seq_id);
  }

  return 0;
}