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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.LinkedList;
import java.util.Queue;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String line = br.readLine();
		String[] arr = line.split(" ");
		int n = Integer.parseInt(arr[0]);
		int q = Integer.parseInt(arr[1]);

		Queue<Process> queue = new LinkedList<Process>();
		for (int i = 0; i < n; i++) {
			line = br.readLine();
			arr = line.split(" ");
			Process p = new Process();
			p.name = arr[0];
			p.time = Integer.parseInt(arr[1]);
			queue.offer(p);
		}

		int passed = 0;
		while (queue.size() > 0) {
			Process p = queue.poll();
			if (p.time <= q) {
				passed += p.time;
				System.out.println(p.name + ' ' + passed);
			} else {
				passed += q;
				p.time -= q;
				queue.offer(p);
			}
		}

		if (br != null) {
			br.close();
			br = null;
		}
	}

	private static class Process {
		public String name;
		public int time;
	}
}