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 Kotlin (1.0.0) (judge verdict: Wrong Answer)


fun main(args: Array<String>) {
    val (a, b) = readLine()!!.split(" ").map { it.toInt() }

    // 最も計算後の絶対値の差分が小さくなる値で変更していけばよいのでは
    // 6通りを N 回試すことになるか

    var now = a
    var cnt = 0

    fun calc(now: Int, diff: Int): Int {
        // +- どちらのほうが近づくか決定
        var next = 0
        if (Math.abs(b - (now + diff)) < Math.abs(b - (now - diff))) {
            return now + diff
        } else {
            return now - diff
        }
    }

    while(now != b) {
        var tmp1 = calc(now, 1)
        var tmp2 = calc(now, 5)
        var tmp3 = calc(now, 10)

        if ( Math.abs(b-tmp1) < Math.abs(b-tmp2)) {
            now = tmp1
        } else {
            now = tmp2
        }

        if ( Math.abs(b - tmp3) < Math.abs(b - now)) {
            now = tmp3
        }

        cnt += 1
    }

    println(cnt+1)
}