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<algorithm>
#include <cassert>
#include<string>
#include<vector>
#include <utility>
#include <math.h>
#include<cmath>
#include<map>
#include <limits>
# include <iomanip>
#define INF 1000000009;
using namespace std;
int gcd(int a, int b) {
	while (a % b != 0) {
		int c = 0;
		c = a % b;
		a = b;
		b = c;
	}
	return  b;
}
vector<int>machi[100009];
vector<pair<int, int> > cope;
long long int copyy[100009];
int manji[100009];
int c = 0;
int main() {
	int N = 0, M = 0,a=0,b=0;
	cin >> N >> M;
	c = max(c, N);
	for (int i = 0; i < M; i++) {
		cin >>a >> b;
		manji[a]++;
		copyy[a]++;
		
		machi[a].push_back(b);
		cope.push_back  (make_pair(a, b));
	}
	//cout << "OK" << endl;
	for (int i = 1; i <= c; i++) {
		if (machi[i].size()>1) {
			sort(machi[i].begin(), machi[i].end());
		}
	}
	//cout << "OK" << endl;
	
	for (long long int i = 0; i < M; i++) {
    auto ans = lower_bound(machi[cope[i].first].begin(), machi[cope[i].first].end(), cope[i].second);
		cout << setfill('0') << right << setw(6) << cope[i].first;
		cout << setfill('0') << right << setw(6) << ans-machi[cope[i].first].begin()+1 << endl;
					
			
		
	}
	return 0;
}