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 Python (3.4.3) (judge verdict: Accepted)

# -*- coding: utf-8 -*-

# from time import time
# 
# start = time()


def six(n):
    n_str = str(n)
    while len(n_str) < 6:
        n_str = "0" + n_str

    return n_str


N, M = map(int, input().split())

id_list = [None] * M
dic_P = {}
info = []
info_idx = 0

for i in range(M):
    P, Y = map(int, input().split())
    if P in dic_P.keys():
        info[dic_P[P]].append([i, P, Y])
    else:
        dic_P[P] = info_idx
        info_idx += 1
        info.append([[i, P, Y]])

for group in info:
    group.sort(key=lambda a_i: a_i[2], reverse=False)
    for i, pair in enumerate(group):
        id = six(pair[1]) + six(i+1)
        id_list[pair[0]] = id

for id in id_list:
    print(id)

# print(time() - start)