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.*;



/*
 * Aizuuuuuuu
 * 
 */
public class Main {
	public static void print(ArrayList<String> list) {
		for (String i :list) {
			System.out.println(i);
		}
	}
	public static void main(String[] args) {
		int globalTime = 0;
		
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int q = sc.nextInt();
		ArrayList<String> results = new ArrayList<String>();
		LinkedList<String> ns = new LinkedList<String>();
		LinkedList<Integer> qs = new LinkedList<Integer>();
		
		while (ns.size() < n) {
			ns.addLast(sc.next());
			qs.addLast(sc.nextInt());
		}
		
		String tempS = "";
		int tempI = 0;
		
		while (ns.size() > 0) {
			if (qs.peek() - q <= 0) {
				results.add(ns.poll() + " " + (qs.peek()+globalTime));
				globalTime += qs.poll();
			}
			else {
				tempI = qs.poll();
				tempI -= q;
				qs.addLast(tempI);
				ns.addLast(ns.poll());
				globalTime += q;
			}
		}
		print(results);
		
	}
}