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, i, j, k;
    //int cnt;
    int n_tmp, y_tmp, i_tmp;
    scanf("%d %d", &N, &M);

    int list[N][M][2]; //0番目,,,入力番号i 1番目,,,year
    int count[N]; //n番目の要素に入っている数
    for( i=0; i<N; i++){
        count[i] = 0;
    }
    for( i=0; i<M; i++){
        scanf("%d %d", &n_tmp, &y_tmp);
        list[n_tmp-1][count[n_tmp-1]][0] = i;
        list[n_tmp-1][count[n_tmp-1]][1] = y_tmp;
        ++count[n_tmp-1];
    }

    for( i=0; i<N; i++){
        //ソート
        for( j=0; j<count[i]; j++){
            for( k=count[i]-1; k > j; k--){
                if( list[i][k][1] < list[i][k-1][1]){
                    i_tmp = list[i][k][0];
                    y_tmp = list[i][k][1];
                    list[i][k][0] = list[i][k-1][0];
                    list[i][k][1] = list[i][k-1][1];
                    list[i][k-1][0] = i_tmp;
                    list[i][k-1][1] = y_tmp;
                }
            }
        }
    }

    //cnt = 0;
    int ans[M][2];
    for( i=0; i<N; i++){
        for( j=0; j<count[i]; j++){
            ans[ list[i][j][0] ][0] = i+1;
            ans[ list[i][j][0] ][1] = j+1;
        }
        //++cnt;
        //if(cnt > M){printf("error\n"); return 1;}
    }

    for( i=0; i<M; i++){
        printf("%06d%06d\n", ans[i][0], ans[i][1]);
    }
    return 0;
}