Problem: Notes

Template in C

Constraints

* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 1000
* 1 ≤ timei ≤ 50000
* 1 ≤ length of namei ≤ 10
* 1 ≤ Sum of timei ≤ 1000000

Input

n q
name1 time1
name2 time2
...
namen timen


In the first line the number of processes n and the quantum q are given separated by a single space.

In the following n lines, names and times for the n processes are given. namei and timei are separated by a single space.

Output

For each process, prints its name and the time the process finished in order.

Example

Input

5 100
p1 150
p2 80
p3 200
p4 350
p5 20


Output

p2 180
p5 400
p1 450
p3 550
p4 800

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
import java.util.concurrent.ArrayBlockingQueue;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        List<Integer> n_q = Stream.of(sc.nextLine().split(" "))
                .map(Integer::parseInt)
                .collect(Collectors.toList());
        int q = n_q.get(1);
        Queue<String> qu = new ArrayBlockingQueue<>(n_q.get(0));
        int t = 0;
        while(sc.hasNext()) {
            qu.add(sc.nextLine());
        }
        while(!qu.isEmpty()) {
            String[] name_time = qu.poll().split(" ");
            int time = Integer.parseInt(name_time[1]);
            if (time > q) {
                time -= q;
                t += q;
                qu.add(name_time[0] + " " + time);
            } else {
                t += time;
                System.out.println(name_time[0] + " " + t);
            }
        }
    }
}