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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

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

#include <bits/stdc++.h>
#include <cmath>
#include <limits>

using namespace std;

map<int, map<long long, int>> mp;
int P[100000+2] = {0};
long long Y[100000+2] = {0};

int main(int argc, char* argv[]){
	int N, M;

	cin >> N >> M;

	for(int i=1; i<=M; i++){
		int p;
		long long y;
		cin >> p >> y;
		P[i] = p; Y[i] = y;
		if(mp.count(p)==0){
			map<long long, int> m;
			m[y] = 1;
			mp[p] = m;
		}
		else{
			mp[p][y] = 1;
		}
	}

	for(int i = 1; i <= M; i++){
		int p = P[i];
		long long y = Y[i];
		int s = (int) distance(mp[p].begin(), mp[p].find(y));
		printf("%06d%06d\n", p, s + 1);
	}
	return 0;
}