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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#define swap(type, x, y) do{ type t = x; x = y; y = t; }while(0)

void quick(int a[], int left, int right, int b[], int c[]){
  int pl = left, pr = right, x = a[(left + right) / 2];
  do{
    while(a[pl] < x)  pl++;
    while(a[pr] > x)  pr--;
    if(pl <= pr){
      swap(int, a[pl], a[pr]);
      swap(int, b[pl], b[pr]);
      swap(int, c[pl], c[pr]);
      pl++, pr--;
    }
  }while(pl <= pr);
  if(pl < right)  quick(a, pl, right, b, c);
  if(pr > left)   quick(a, left, pr, b, c);
}

int main(){
  int n, m;  scanf("%d %d", &n, &m);
  int p[100005], y[100005], ad[100005];
  for(int i=1; i<=m; i++){
    scanf("%d %d", p+i, y+i);
    ad[i] = i;
  }
 
  quick(y, 1, m, p, ad);

  char ans[100005][14];
  int num[100005] = {0};
  
  for(int i=1; i<=m; i++){
    int pn = p[i], bn = ++num[p[i]];
    int d = 1;
    for(int j=0; j<6; j++){
      ans[ad[i]][5-j]  = (char)((int)'0'+(pn/d)%10);
      ans[ad[i]][11-j] = (char)((int)'0'+(bn/d)%10);
      d *= 10;
    }
  }

  for(int i=1; i<=m; i++){
    for(int j=0; j<12; j++)
      printf("%c", ans[i][j]);
    puts("");
  }

  return 0;
}