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: Wrong Answer)

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

using ll = long long int;

const int MAX_N = (int)(1e5 + 5);
const int MAX_M = (int)(1e5 + 5);

int n, m;

map<int, vector<pair<ll, ll>>> cnt;

map<int, pair<int, int>> ans;

using P = pair<ll, int>;

int main(void) {
    // Here your code !
    scanf("%d %d", &n, &m);

    for (int i = 1; i <= m; ++i) {
      ll p;
      ll y;
      scanf("%lld %lld", &p, &y);

      cnt[p].push_back(make_pair(y, i));
    }

    for (auto &e: cnt) {
      auto tmpvec = e.second;
      sort(tmpvec.begin(), tmpvec.end(), [&](P a, P b) { return a.first < b.first; } );
      for (int i = 0; i < tmpvec.size(); ++i) {
        auto obj = tmpvec[i];
        ans[obj.second] = make_pair(e.first, i + 1);
      }
    }

    for (auto &e: ans) {
      int fst = 6 - (e.second.first / 10) - 1;
      fst = (fst > 6) ? 6 : fst;
      for (int i = 0; i < fst; ++i) {
        printf("%d", 0);
      }
      printf("%d", e.second.first);

      int snd = 6 - (e.second.second / 10) - 1;
      snd = (snd > 6) ? 6 : snd;
      for (int i = 0; i < snd; ++i) {
        printf("%d", 0);
      }
      printf("%d", e.second.second);

      printf("\n");
    }

    return 0;
}