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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, M;
  scanf("%d %d", &N, &M);
  int P[M], Y[M], X[M];
  int I[N][M] = {};
  for(int m=0; m<M; m++){
    scanf("%d %d", &P[m], &Y[m]);
    X[m] = 1;
    for(int i=0; i<=m; i++){
      if(I[P[m]-1][i] == 0){
        I[P[m]-1][i] = m + 1;
        break;
      }
      if(Y[I[P[m]-1][i]-1] > Y[m]) X[I[P[m]-1][i]-1]++;
      else X[m]++;
    }
  }
  
  for(int m=0; m<M; m++)printf("%06d%06d\n", P[m], X[m]);
  return 0;
}