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) {
		
		int numberOfProcesses;
		int quantum;
		int clock = 0;
		String line;
		
		Queue<String> taskQueue = new LinkedList<String>();
		Queue<Integer> timeQueue = new LinkedList<Integer>();
		
		Scanner keyboard = new Scanner(System.in);
		Scanner input;
		
		line = keyboard.nextLine();
		
		input = new Scanner(line);
		
		numberOfProcesses = Integer.parseInt(input.next());
		quantum = Integer.parseInt(input.next());
		
		for(int i = 0; i < numberOfProcesses; i++) {
			line = keyboard.nextLine();
			input = new Scanner(line);
			
			taskQueue.add(input.next());
			timeQueue.add(Integer.parseInt(input.next()));
		}
		
		while(!taskQueue.isEmpty()) {
			
			String currentTask = taskQueue.remove();
			int currentTime = timeQueue.remove();
			
			if(currentTime > quantum) {
				currentTime = currentTime - quantum;
				
				taskQueue.add(currentTask);
				timeQueue.add(currentTime);
				
				clock = clock + quantum;
			}
			else {
				
				clock = clock + currentTime;
				
				System.out.print(currentTask + " ");
				System.out.println(clock);
			}
		}
		
		
		
	}

}