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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <cstdlib>
#include <math.h>
#include <algorithm>
#include <numeric>
using namespace std;

int main(){
  int N, M, tP, tY;
  vector<int> P, Y, rank, flag;
  cin >> N >> M;
  for (int i=0; i < M; i++) {
    scanf("%d%d", &tP, &tY);
    P.push_back(tP);
    Y.push_back(tY);
    rank.push_back(0);
    flag.push_back(0);
  }
  for (int i=0; i < M; i++) {
    for (int j=0; j < M; j++) {
      if(flag[i] == 1) continue;
      if(P[i] == P[j]) {
        if (Y[i] < Y[j]) {
          rank[j]++;
        } else {
          rank[i]++;
        }
        flag[j] = 1;
      }
    }
  }
  for (int i=0; i < M; i++) {
    printf("%06d", P[i]);
    printf("%06d\n", rank[i]);
  }
  return 0;
}