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(void){

    int n,m;

    scanf("%d",&n);
    scanf("%d",&m);

    int p[100000],y[100000];
    int num[100000][100000];
    int count[100000];
    int counta,countb;

    //ok
    for(counta=0;counta<m;counta++){
       scanf("%d",&p[counta]);
       scanf("%d",&y[counta]);
    }
    //ok
    for(counta=0;counta<n;counta++){
        count[counta]=0;
    }
    //ok
    for(counta=0;counta<m;counta++){
        num[p[counta]-1][count[p[counta]-1]]=y[counta];
        count[p[counta]-1]++;
    }

    for(counta=0;counta<m;counta++){
        //ok
        printf("%06d",p[counta]);

        int countd=0;

        for(countb=0;countb<count[p[counta]-1];countb++){
            if(y[counta]>num[p[counta]-1][countb]){
                countd++;
            }
        }
        printf("%06d\n",countd+1);

    }
    return 0;
}