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()
n=n.to_i
m=m.to_i
p=[]
y=[]
y_so=[]
ans=[]
cou=1
b_num=0
m.times do |i|
  p[i],y[i]=gets.split()
  p[i]=p[i].to_i
  y[i]=y[i].to_i
end
y_so=y.sort
m.times do |i|
  num=y.index(y_so[i])
  ans[num]=ans[num].to_s
  (6-p[num].to_s.length).times{ans[num]=ans[num]+"0"}
  ans[num]=ans[num]+p[num].to_s
  if i==0||p[b_num]!=p[num] then
    cou=1
  else 
    cou+=1
  end
  (6-cou.to_s.length).times{ans[num]=ans[num]+"0"}
  ans[num]=ans[num]+cou.to_s
  b_num=num
end
m.times do |i|
  puts ans[i]
end