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 <iostream>
#include <string>
#include <map>
#include <algorithm>
#include <vector>

using namespace std;

int main(void){
    int n, m;
    vector<pair<int, pair<int, int>>> py;
    int f = -1;
    int b_count;
    vector<pair<int, string>> ans;
    

    cin >> n >> m;
    for(int i = 0; i < m; i++){
        int p, y;
        cin >> p >> y;
        py.push_back(make_pair(p,make_pair(y, i)));
    }

    /*
    for(int i = 0; i < m; i++){
        cout << py[i].first << " " << py[i].second.first << " " << py[i].second.second << endl;
    }
    */

    sort(py.begin(), py.end());

    /*
    for(int i = 0; i < m; i++){
        cout << py[i].first << " " << py[i].second.first << " " << py[i].second.second << endl;
    }
    */

    for(int i = 0; i < m; i++){
        if(f != py[i].first){
            f = py[i].first;
            b_count = 1;
        }
        else{
            b_count++;
        }
        int a_num = to_string(py[i].first).length();
        int b_num = to_string(b_count).length();
        
        ans.push_back(make_pair(py[i].second.second, string(6-a_num, '0') + to_string(py[i].first) + string(6-b_num, '0') + to_string(b_count)));
    }

    sort(ans.begin(), ans.end());

    for(int i = 0; i < m; i++){
        cout << ans[i].second << endl;
    }
    return 0;
}