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: Wrong Answer)

#include <stdio.h>

void q_sort(int numbers[], int index[], int left, int right)
{
    int pivot, ipivot, l_hold, r_hold;

    l_hold = left;
    r_hold = right;
    pivot = numbers[left];
    ipivot = index[left];
    while (left < right)
    {
        while ((numbers[right] >= pivot) && (left < right))
            right--;
        if (left != right)
        {
            numbers[left] = numbers[right];
            index[left] = index[right];
            left++;
        }
        while ((numbers[left] <= pivot) && (left < right))
            left++;
        if (left != right)
        {
            numbers[right] = numbers[left];
            index[right] = index[left];
            right--;
        }
    }
    numbers[left] = pivot;
    index[left] = ipivot;
    pivot = left;
    left = l_hold;
    right = r_hold;
    if (left < pivot)
        q_sort(numbers, index, left, pivot-1);
    if (right > pivot)
        q_sort(numbers, index, pivot+1, right);
}

int main(void){
    // Your code here!
    int N,M,P[100000],Y[100000],index[100000],stone[100000],no[100000];
    scanf("%d %d",&N,&M);
    for(int i = 0; i < M; i++)
    {
        scanf("%d %d", &P[i], &Y[i]);
        index[i] = i;
        stone[i] = 1;
    }
    q_sort(Y,index,0,M-1);
    for(int i = 0; i < M; i++)
    {
        no[i] = stone[P[index[i]]-1]++;
    }
    q_sort(index,no,0,M-1);
    for(int i = 0; i < M; i++)
    {
        printf("%06d%06d\n",P[i],no[i]);
    }
}