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.Deque;
import java.util.Scanner;

class Processees {
	private String name;
	private int time;

	public String getName() {
		return name;
	}

	public int getTime() {
		return time;
	};

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

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

}


public class Main {

	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int p = scanner.nextInt();

		Deque<Processees> Processeeses = new ArrayDeque<Processees>(n);
		for (int i = 0; i < n; i++) {
			Processeeses.add(new Processees(scanner.next(), scanner.nextInt()));
		}
		int totalTime =0;
		int tempTime;
		while(Processeeses.size()!=0){
			if(Processeeses.element().getTime()<=p){
				totalTime+=Processeeses.element().getTime();
				System.out.println(Processeeses.element().getName()
						+ " "+totalTime);
				Processeeses.poll();
			}
			else{
				tempTime = Processeeses.element().getTime()-p;
				Processeeses.element().setTime(tempTime);
				Processeeses.add(Processeeses.element());
				Processeeses.poll();
				totalTime+=p;
			}
		}

	}
}