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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n,m = gets.split.map(&:to_i)
ary = []
rank = []
m.times do |i|
  ary[i] = gets.split.map(&:to_i)
end

(1..n).each do |i|
  rank[i] = []
end
m.times do |i|
  p,y = ary[i]
  rank[p].push([y,i]) 
end

(1..n).each do |p|
  next if rank[p].nil?
  rank[p].sort!
end

(1..n).each do |p|
  next if rank[p].nil?
  (rank[p].size).times do |r|
    ary[rank[p][r][1]][2] = r+1
  end
end
p rank
m.times do |i|
  puts ary[i][0].to_s.rjust(6,"0") + ary[i][2].to_s.rjust(6,"0")
end