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; 
import java.util.Queue; 

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int ts = 0;
		
		Queue<Integer> qN = new LinkedList<>(); 
		Queue<String> qP = new LinkedList<>(); 
		
		String line = sc.nextLine();
		
		String[] lineList = line.split(" ");
		String[] itemsList = new String[Integer.parseInt(lineList[0])*2];
		
		int q = Integer.parseInt(lineList[1]);
		
		String[] items = new String[Integer.parseInt(lineList[0])];
		
		//add everything into queues
		for(int i = 0; i < Integer.parseInt(lineList[0]); i++) {
			items[i] = sc.nextLine();
			itemsList = items[i].split(" ");
			qP.add(itemsList[0]);
			qN.add(Integer.parseInt(itemsList[1]));
		}
		
		while(!qP.isEmpty()){
			if(q - qN.peek() < 0) {
				qN.add(-1*(q-qN.remove()));
				qP.add(qP.remove());
				ts += q;
			}
			else {
				ts += qN.remove();
				System.out.println(qP.remove() + " " + ts);
			}
		}
		
	}
}