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.Scanner;

public class Main {

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

		LinkedList<Process> processQueue = new LinkedList<Process>();

		int n = sc.nextInt();
		int quantum = sc.nextInt();
		for (int i = 0; i < n; i++) {
			processQueue.add(new Process(sc.next(), sc.nextInt()));
		}
		
		for (int spendTime = 0; !processQueue.isEmpty(); ) {
			Process run = processQueue.remove();
			if (run.pRemain > quantum) {
				run.pRemain -= quantum;
				processQueue.add(run);
				spendTime += quantum;
			} else if (run.pRemain == quantum) {
				run.pRemain -= quantum;
				spendTime += quantum;
				printProcess(run, spendTime);
			} else {
				spendTime += run.pRemain;
				run.pRemain = 0;
				printProcess(run, spendTime);
			}
		}
    	
        sc.close();
	}

	private static void printProcess(Process run, int eTime) {
		StringBuffer str = new StringBuffer();

		str.append(run.pName);
		str.append(" ");
		str.append(eTime);

		System.out.println(str);
	}
}

class Process {
	String pName;
	int pRemain;
	int eTime;
	
	Process(String name, int time) {
		pName = name;
		pRemain = time;
		eTime = 0;
	}
}