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: Time Limit Exceeded)

N,M=map(int,input().split())
l1=list()
for i in range(M):
   l1x=list(map(int,input().split()))
   l1.append(l1x)
l2=l1.copy()
l2.sort()
x=1
l31="{0:06d}".format(l2[0][0])+"{0:06d}".format(x)
l3=[l31]
for i in range(1,M):
   if l2[i-1][0]==l2[i][0]:
      x+=1
      l3x="{0:06d}".format(l2[i][0])+"{0:06d}".format(x)
      l3.append(l3x)
      
   else:
      x=1
      l3x="{0:06d}".format(l2[i][0])+"{0:06d}".format(x)
      l3.append(l3x)

for i in range(M):
   li=l2.index(l1[i])
   print(l3[li])