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 <iostream>
#include <cstdlib>
#include <queue>
#include <iomanip>
using namespace std;
int main(){
	int n, m;
	priority_queue<pair<long int, int>, vector<pair<long int, int> >, greater<pair<long int, int> > > que[100000];
	cin >> n >> m;
	int p, y;
	for(int i = 0;i < m;i++){
		cin >> p >> y;
		que[p - 1].push(make_pair(y, i));
	}
	long long int number[100000];
	int count;
	pair<long int, int> e;
	for(int i = 0;i < n;i++){
		count = 1;
		while(que[i].size() > 0){
			e = que[i].top();
			que[i].pop();
			number[e.second] = (i + 1) * 1000000 + count;
			count++;
		}
	}
	for(int i = 0;i < m;i++){
		cout << setfill('0') << setw(12) << right << number[i] << endl;
	}
}