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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>

int main()
{
        int N, M;
        int i, j, rank=0, ymax=0;
        int P[10000]={}, Y[10000]={};
        int RNK[10000]={};
        int max=0;

        scanf("%d %d", &N, &M);
        for(i=0; i<M; i++)
        {
                scanf("%d %d", &P[i], &Y[i]);
                if(max<P[i])max=P[i];
        }
        for(i=1; i<=max; i++, rank=0)
        {
                for(j=0; j<M; j++)
                {
                        if(P[i]==P[j]&&Y[j]>ymax)
                        {
                                ymax = Y[j];
                                rank = rank + 1;
                        }
                        RNK[j]=rank;
                }
                ymax = 0;
        }
        for(i=0; i<M; i++)
        {
                printf("%06d", P[i]);
                printf("%06d\n", RNK[i]);
        }
        return 0;
}