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.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main{

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        String[] row = scan.nextLine().split(" ");

        int n = Integer.parseInt(row[0]);
        int q = Integer.parseInt(row[1]);

        Queue<Process> que = new LinkedList<Process>();
        for (int i=0; i<n; i++) {
            String[] strArray = scan.nextLine().split(" ");

            Process process = new Process(strArray[0],
					  Integer.parseInt(strArray[1]));
            que.offer(process);
        }

        int resultTime = 0;
        while (que.size() > 0) {
            Process process = que.poll();
            resultTime += q;
            if ((process.processTime = process.processTime - q) > 0) {

                que.offer(process);
                continue;
            }
            resultTime += process.processTime;

            System.out.println(process.name + " " + resultTime);
        }

    }

    private final static class Process {
        public String name = null;
        public int processTime = 0;

        public Process(String name, int processTime) {
            this.name = name;
            this.processTime = processTime;
        }
    }
}