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.*;
import static java.lang.System.*;

public class Main {
	static Scanner sc = new Scanner(System.in);
	public static void main(String[] args) {
		int n = sc.nextInt();
		int q = sc.nextInt();
		Deque<Process> deque = new ArrayDeque<>();
		String s;
		int a;
		for (int i=0; i<n; i++) {
			s = sc.next();
			a = sc.nextInt();
			deque.addLast(new Process(s,a));
		}
		int time = 0;
		int temp;
		while (deque.size()!=0) {
			temp = Math.min(q,deque.getFirst().getTime());
			deque.getFirst().time -= temp;
			time += temp;
			if (deque.getFirst().getTime() > 0) {
				deque.addLast(deque.removeFirst());
			}
			else {out.println(deque.removeFirst().getName()+" "+time);}
		}
	}
}

class Process {
	String name;
	int time;
	public Process (String n, int t) {
		this.name = n;
		this.time = t;
	}
	public String getName () {
		return this.name;
	}
	public int getTime () {
		return this.time;
	}
}