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.Queue;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {

		Scanner cin = new Scanner(System.in);
		while (cin.hasNext()) {
			String[] str = cin.nextLine().split(" ");
			int rundex = Integer.parseInt(str[0]);
			int step = Integer.parseInt(str[1]);
			Queue<node> qu = new LinkedList<node>();
			int time =0;
			for (int i = 1; i <= rundex; i++) {
				str = cin.nextLine().split(" ");
				qu.add(new node(str[0], Integer.parseInt(str[1]),0));
			}
			while (!qu.isEmpty()) {
				node temp = qu.poll();
				if (temp.value <= step) {
					time+=temp.value;
					System.out.println(temp.point +" "+time);
				}
				else{
					time+=step;
					temp.value-=step;
					qu.add(temp);
				}
			}
		}
	}

	static class node {
		String point;
		int value;
		int usetime;
		public node(String point, int value,int usetime) {
			this.point = point;
			this.value = value;
			this.usetime=usetime;
		}
	}
}