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.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        String[] values = scanner.nextLine().split(" ");
        int n = Integer.parseInt(values[0]);
        int limit = Integer.parseInt(values[1]);

        Queue<Process> queue = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            queue.add(new Process(scanner.nextLine().split(" ")));
        }

        int elapsedTime = 0;
        while (!queue.isEmpty()) {
            Process process = queue.poll();
            int processTime = Math.min(process.reamingTime, limit);
            process.reamingTime -= processTime;
            elapsedTime += processTime;
            if (process.reamingTime == 0) {
                System.out.println(process.processName + " " + Integer.toString(elapsedTime));
            } else {
                queue.add(process);
            }
        }

    }

    private static class Process {
        String processName;
        int reamingTime;

        Process(String[] data) {
            processName = data[0];
            reamingTime = Integer.parseInt(data[1]);
        }

    }
}