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.*;
import java.util.LinkedList;
import java.util.Queue;

class Pair{
	public String key;
	public int value;
	public Pair(String name, int number) {
		this.key = name;
		this.value = number;
	}
}

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String[] info = sc.nextLine().split(" ");
		
		int n = Integer.parseInt(info[0]);
		int limit =Integer.parseInt(info[1]);
		int time = 0;
		
		Queue<Pair> q = new LinkedList<Pair>();
		String [] inputArrayS = new String[n];
		for (int i = 0; i < n; i += 1) {
			inputArrayS[i]  = sc.nextLine();
		}
		sc.close();
		
		for (int j = 0; j < n; j += 1) {
			
			int quantum = Integer.parseInt(inputArrayS[j].split(" ")[1]);
			String nameS = inputArrayS[j].split(" ")[0];
			Pair p = new Pair(nameS, quantum);
			q.offer(p);
			}

			while (q.size()>0) {
				
					if(q.peek().value > limit) {
						Pair cp = q.remove();
						time = time + limit;
						
					
					cp.value = cp.value - limit;
					cp.key = cp.key;	
					q.add(cp);
					
					
				}else{
					
					time = time + q.peek().value;				
					System.out.println(q.peek().key + " " + time);
					q.remove();
					
				}
			}
		}
	}