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;

public class Main {
	public static int head1 = 0;
	public static int head2 = 0;
	public static int tail = 0;
	public static String[] queue1 = new String[1000000];
	public static int[] queue2 = new int[1000000];
	
	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		
		int n = stdIn.nextInt();
		int q = stdIn.nextInt();
		
		for (int i = 0; i < n; i++) {
			String a = stdIn.next();
			int b = stdIn.nextInt();
			enqueue(a, b);
		}
		
		int time = 0;
		while (head1 != tail) {
			String a = strDequeue();
			int b = intDequeue();
			if (b <= q) {
				time += b;
				System.out.println(a +" "+ time);
			}
			else {
				time += q;
				b -= q;
				enqueue(a, b);
			}
		}
	}
	
	public static void enqueue(String x, int y) {
		queue1[tail] = x;
		queue2[tail] = y;
		tail++;
	}
	
	public static String strDequeue() {
		return queue1[head1++];
	}
	
	public static int intDequeue() {
		return queue2[head2++];
	}
}