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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

console.log((args=>{
  const [[N,M],...tmp] = args.trim().split('\n').map( r=>r.split(' ').map(v=>v|0) );
  const p = Array.from({length:N},_=>[]);
  for ( let i = 0; i < M; i++ ) {
    const [P,Y] = tmp[i];
    p[P-1].push([i,Y]);
  }
  p.forEach(r=>r.sort((a,b)=>a[1]-b[1]));
  const a = Array.from(tmp,([p])=>`${p}`.padStart(6,0));
  p.forEach(r=>r.forEach(([i],j)=>a[i] += `${j+1}`.padStart(6,0)))
  return a.join`\n`;
})(require('fs').readFileSync('/dev/stdin', 'utf8')));