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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main(){
  int M,N;
  cin>>M>>N;
  vector<pair<pair<int,int>,int>> A(N);
  for(int i=0;i<N;i++){
    cin>>A[i].first.second>>A[i].first.first;
    A[i].second=i;
  }
  sort(A.begin(),A.end());
  vector<int> B(M+1,1),C(N),D(N);
  for(int i=0;i<N;i++){
    C[A[i].second]=B[A[i].first.second];
    B[A[i].first.second]++;
    D[A[i].second]=A[i].first.second;
  }
  for(int i=0;i<N;i++){
    for(int j=to_string(D[i]).size();j<6;j++)
      cout<<0;
    cout<<D[i];
    for(int j=to_string(C[i]).size();j<6;j++)
      cout<<0;
    cout<<C[i]<<endl;
  }
}