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 sc = new Scanner(System.in);

		String[] S = sc.nextLine().split(" ");
		int n = Integer.parseInt(S[0]);
		int quantum = Integer.parseInt(S[1]);

		Queue<Node> queue = new ArrayDeque<Node>();

		for (int i = 0; i < n; i++) {
			S = sc.nextLine().split(" ");
			Node node = new Node(S[0], Integer.parseInt(S[1]));
			queue.offer(node);
		}

		long count = 0;

		while (!queue.isEmpty()) {
			Node tmp = queue.poll();
			if (tmp.getTime() > quantum) {
				count += quantum;
				tmp.setTime(tmp.getTime() - quantum);
				queue.offer(tmp);
			} else {
				count += tmp.getTime();
				System.out.printf("%s %d", tmp.getName(), count);
				System.out.println("");
			}
		}

		sc.close();
	}
}

class Node {
	private String name;
	private int time;

	public Node(String name, int time) {
		this.name = name;
		this.time = time;
	}

	public void setName(String name) {
		this.name = name;
	}

	public String getName() {
		return this.name;
	}

	public void setTime(int time) {
		this.time = time;
	}

	public int getTime() {
		return this.time;
	}
}