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 "bits/stdc++.h"
using namespace std;

int main() {
	long long N, M, P, Y, B[100000], count = 1, NOW = 1;
	static pair<long long, int> A[100000];
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		cin >> P >> Y;
		A[i].first = P * pow(10, 9) + Y, A[i].second = i;
	}
	sort(A, A + M);
	for (int i = 0; i < M; i++) {
		long long F = A[i].first / 1000000000, S = A[i].first % 1000000000;
		if (NOW == F) {
			B[A[i].second] = NOW * 1000000 + count;
			count++;
		}
		else {
			count = 1;
			NOW = F;
			B[A[i].second] = NOW * 1000000 + count;
			count++;
		}
	}
	for (int i = 0; i < M; i++) {
		cout << setfill('0') << setw(12) << B[i] << endl;
	}
}