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<vector>
#include<algorithm>
#include<cmath>
#include<iomanip>
using namespace std;

int main(){
  int n , m;
  cin >> n >> m;
  vector< vector<long long> > a(n);
  vector< pair<int ,long long> >b;
  for(int i = 0; i < m; i++){
    long long p, y;
    cin >> p >> y;
    a[p-1].push_back(y);
    b.push_back(make_pair(p,y));
  }
  for(int i = 0; i < n; i++){
    sort(a[i].begin(), a[i].end());
  }
  for(int i = 0; i < m; i++){
    for(int j = 0; j < m; j++){
      if(a[b[i].first-1][j] == b[i].second){
        cout << setfill('0') << setw(6) << right << b[i].first;
        cout << setfill('0') << setw(6) << right << j+1 << endl;
        break;
      }
    }
  }
}