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 LuaJIT (2.0.4) (judge verdict: Accepted)

local ior = io.input()
local n, m = ior:read("*n", "*n")
local tbl = {}
for i = 1, m do
  local p, y = ior:read("*n", "*n")
  table.insert(tbl, {i, p, y})
end
table.sort(tbl, function(x, y)
  if x[2] ~= y[2] then
    return x[2] < y[2]
  else
    return x[3] < y[3]
  end
end
)

local curp = 0
local cury = 0
for i = 1, m do
  if curp ~= tbl[i][2] then
    cury = 0
    curp = tbl[i][2]
  end
  cury = cury + 1
  tbl[i][3] = cury
end

table.sort(tbl, function(x, y)
  return x[1] < y[1]
end
)

for i = 1, m do
  io.write(string.format("%06d", tbl[i][2]))
  io.write(string.format("%06d", tbl[i][3]))
  io.write("\n")
end