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: Runtime Error)

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

import math

N, M = map(int, input().split())
P = [None] * M
Y = [None] * M
for i in range(M):
    p, y = map(int, input().split())
    P[i] = p
    Y[i] = y

ken = [None] * N
for i in range(N):
    ken[i] = []

for i in range(M):
    ken[P[i] - 1].append(Y[i])


# < M * log(N)
for i in range(N):
    ken[i] = sorted(ken[i])


def close_twopower(n):
    output = 2
    while output < n:
        output = output * 2
    return output


def comp(tr, node, x, i):
    if tr[i] is None:
        return i
    elif x < tr[i][0]:
        return comp(tr, node, x, 2 * i + 1)
    else:
        return comp(tr, node, x, 2 * i + 2)


# 木構造
def create_bitree(a):
    # a = [[int, hoge, ...], ...]
    n = len(a)
    output = [None] * int(math.pow(2, n))

    for i in range(n):
        # output = comp(output, a[i], a[i][0], 0)
        output[comp(output, a[i], a[i][0], 0)] = a[i]

    return output


city_info = [None] * M
k = 0

# count M
while(k < M):
    for l in range(N):
        for m in range(len(ken[l])):
            # ken[l][m]の値によって二分木を作る
            # create_bitree()

            city_info[k] = [ken[l][m], l + 1, m + 1]
            k += 1

bi_city_info = create_bitree(city_info)


def create_id(p, x):
    output_p = str(p)
    output_x = str(x)
    p_len = len(output_p)
    x_len = len(output_x)
    while p_len < 6:
        output_p = "0" + output_p
        p_len += 1
    while x_len < 6:
        output_x = "0" + output_x
        x_len += 1

    return output_p + output_x


def comp2(tr, x, i):
    if tr[i][0] == x:
        return i
    elif x < tr[i][0]:
        return comp2(tr, x, 2 * i + 1)
    else:
        return comp2(tr, x, 2 * i + 2)

    return 0  # dummy


def bi_search(tr, x):
    # tr is [[int. hoge, ...]
    return comp2(tr, x, 0)


# 律速
# M^2
for i in range(M):
    # Y[i]を探してきてprint
    j = bi_search(bi_city_info, Y[i])
    print(create_id(bi_city_info[j][1], bi_city_info[j][2]))