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.LinkedList;

public class Main {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int nProcesses = input.nextInt();
		int quantum = input.nextInt();
		input.nextLine(); // Clear \n.
		
		LinkedList<Integer> queue = new LinkedList<>();
		
		String[] processes = new String[nProcesses];
		int[] times = new int[nProcesses];
		
		for (int i = 0; i < nProcesses; i++) {
			processes[i] = input.next();
			times[i] = input.nextInt();
			queue.add(i); // Add process to queue.
			input.nextLine(); // Clear \n.
		}

		int elapsedTime = 0;
		while (!queue.isEmpty()) {
			int processID = queue.removeFirst();
			int timeSpent = Math.min(times[processID], quantum);
			times[processID] -= timeSpent;
			elapsedTime += timeSpent;
			
			if (times[processID] == 0) {
				System.out.println(processes[processID] + " " + elapsedTime);
			} else {
				queue.addLast(processID);
			}
		}
		
	}	
}