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: Accepted)

#include <bits/stdc++.h>

using namespace std;
using pii = pair<int, int>;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, M;
  cin >> N >> M;
  vector<vector<pii>> PY(N);
  for (int i = 0; i < M; i++) {
    int p, y;
    cin >> p >> y;
    PY[--p].push_back(make_pair(y, i));
  }
  vector<pii> res(M);
  for (int i = 0; i < N; i++) {
    if (PY[i].empty()) continue;
    sort(PY[i].begin(), PY[i].end());
    for (int j = 0; j < PY[i].size(); j++)
      res[PY[i][j].second] = make_pair(i, j + 1);
  }
  for (int i = 0; i < M; i++) {
    cout << setfill('0') << right << setw(6) << res[i].first + 1;
    cout << setfill('0') << right << setw(6) << res[i].second;
    cout << '\n';
  }
  return 0;
}