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 <bits/stdc++.h>
using namespace std;

int countDigits(int n) {
	int i = 1;
	while (n > 9) {
		n /= 10;
		i++;
	}
	return i;
}

string make6(int n) {
	string s;
	int digits = countDigits(n);
	for (int i = 0; i < 6 - digits; i++) {
		s += '0';
	}
	s += to_string(n);
	return s;
}

int getNumber(vector<int> data, int left, int right, int target) {
	int mid = (left + right) / 2;
	while(true) {
		mid = (left + right) / 2;
		if (data.at(mid) == target) {
			return mid;
		}
		else if (data.at(mid) < target) {
			left = mid + 1;
		}
		else {
			right = mid - 1;
		}
	}
} 

int main(){
    int n,m;
	cin >> n >> m;
	vector<int> p(m),y(m);
	vector<vector<int>> group((n + 1), vector<int>(0, 0));
	for (int i = 0; i < m; i++) {
		cin >> p.at(i) >> y.at(i);
		group.at(p.at(i)).push_back(y.at(i));
	}
    
	for (int i = 1; i <= n; i++) {
		sort(group.at(i).begin(),group.at(i).end());
	}
	
	vector<string> kami6(n + 1);
	for (int i = 0; i < m; i++) {
		kami6.at(p.at(i)) = make6(p.at(i));
	}
	
	vector<string>shimo6(m);
	int number;
	for (int i = 0; i < m; i++) {
		number = getNumber(group.at(p.at(i)), 0, group.at(p.at(i)).size() - 1, y.at(i)) + 1;
		shimo6.at(i) = make6(number);
	}
	
	for (int i = 0; i < m; i++) {
		cout << kami6.at(p.at(i)) + shimo6.at(i) << endl;
	}
}