Problem:
There are n processes in a queue. Each process has namei and timei. The round-robin scheduling handles the processes in order. A round-robin scheduler gives each process a quantum (a time slot) and interrupts the process if it is not completed by then. The process is resumed and moved to the end of the queue, then the scheduler handles the next process in the queue.

For example, we have the following queue with the quantum of 100ms.

A(150) - B(80) - C(200) - D(200)

First, process A is handled for 100ms, then the process is moved to the end of the queue with the remaining time (50ms).

B(80) - C(200) - D(200) - A(50)

Next, process B is handled for 80ms. The process is completed with the time stamp of 180ms and removed from the queue.

C(200) - D(200) - A(50)

Your task is to write a program which simulates the round-robin scheduling.

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.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 1000

1 ≤ timei ≤ 50000

1 ≤ length of namei ≤ 10

1 ≤ Sum of timei ≤ 1000000

Sample Input 1

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

Sample Output 1

p2 180
p5 400
p1 450
p3 550
p4 800

Notes

Template in C

Solution in JAVA (judge verdict: Accepted)

import java.io.IOException;
import java.util.Scanner;
public class Main {

	private static int max; // 要素の数
	private static int run; // 処理速度
	private static Object q[]; // queue
	private static String end[]; // 処理済み
	private static int count; // 処理済み数
	private static int front; // 先頭
	private static int rear; // 末尾
	private static int time; // 合計処理時間

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

		max = sc.nextInt();
		run = sc.nextInt();

		q = new Object[max+1];
		front = rear = count = time = 0;
		end = new String[max];

		for(int i=0;i<max;i++) {
			String a = sc.next();
			int b = sc.nextInt();
			enqueue(a + " " +b);
		}

		while(true) {
			String[] line = dequeue().toString().split(" ");
			int x = Integer.parseInt(line[1]);
			// 処理時間以内なら
			if(x <= run) {
				time += x;
				end[count++] = line[0] + " " + Integer.toString(time);
			// 処理時間以上なら
			}else {
				line[1] = Integer.toString(x - run);
				enqueue(line[0] + " " + line[1]);
				time += run;
			}

			if(count == max) {
				break;
			}
		}

		show();

		sc.close();
	}

	public static void show() {
		for(int i=0;i<max;i++) {
			System.out.println(end[i]);
		}
	}

	/*
	 * ポインタの位置
	 */
	public static int next(int a) {
		return (a + 1) % max;
	}

	/*
	 * 入れる
	 */
	public static void enqueue(Object x) {
/*		if(next(rear) == front) {
			throw new IllegalStateException();
		}*/
		q[rear] = x;
		rear = next(rear);
	}

	/*
	 * 取り出す
	 */
	public static Object dequeue() {
		Object x = q[front];
		q[front] = null;
		front = next(front);
		return x;
	}
}