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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

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

#define PMAX 100000
#define MMAX 100000

typedef struct _city {
	int p_idx;
	int no;
	long y;
} CITY;

int main (void) {
	int n, m;
	CITY cities[MMAX];

	scanf ("%d %d", &n, &m);
	for (int i=0; i<m; i++) {
		scanf ("%d %ld", &cities[i].p_idx, &cities[i].y);
		int no = 1;
		for (int j=0; j<i; j++) {
			if (cities[j].p_idx==cities[i].p_idx) {
				if (cities[j].y > cities[i].y)
					cities[j].no ++;
				else
					no ++;
			}
		}
		cities[i].no = no;
	}

	for (int i=0; i<m; i++)
		printf ("%06d%06d\n", cities[i].p_idx, cities[i].no);

	return 0;
}