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>

#define rep(i, n) for(int i = 0; i < n; ++i)
typedef long long ll;
using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> P(n + 1, vector<int>());
    vector<pair<int, int>> city;
    int p, y;
    rep(i, m){
        cin >> p >> y;
        P[p].push_back(y);
        city.push_back(make_pair(p, y));
    }
    rep(i, n) sort(P[i].begin(), P[i].end());
    rep(i, m){
        vector<int>::iterator iter = find(P[city[i].first].begin(), P[city[i].first].end(), city[i].second);
        size_t idx = distance(P[city[i].first].begin(), iter);
        cout << setfill('0') << setw(6) << right << to_string(city[i].first) << setfill('0') << setw(6) << right << to_string(idx + 1) << '\n';
    }
}