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

public class Main{
	private static int timer = 0;

	public static void processor(Queue<Process> p, int q){
		Process x;
		timer = 0;
		while(!p.isEmpty()){
			x = p.remove();
			if(!x.handle(q)) p.add(x);
		}
	}	

	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int quantum = sc.nextInt();

		Queue<Process> p = new LinkedList<Process>();
		String name;
		int time;
		for(int i=0; i<n; i++){
			name = sc.next();
			time = sc.nextInt();
			Process x = new Process(name, time);
			p.add(x);
		}

		sc.close();

		processor(p,quantum);
	}

	private static class Process{
		String name;
		int time; //remaining time

		public Process(String n, int t){
			name = n;
			time = t;
		}

		//return if it ends
		public boolean handle(int q){
			if(q<time){
				time -= q;
				timer +=q;
				return false;
			} 
			else{
				timer += time;
				System.out.println(name+" "+timer);
				return true;
			}
		}
	}
}