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.*;

public class Main {
	public static void main(String[] args) {
		ALDS1_3_B();
	}

	public static void ALDS1_3_B() {

		class Pair{
			private String name;
			private int time;
			
			public Pair(String name, int time) {
				this.name = name;
				this.time = time;
			}

			String getName(){
				return this.name;
			}

			int getTime(){
				return this.time;
			}
		}

		Scanner sc = new Scanner(System.in);
		Queue<Pair> que = new LinkedList<Pair>();
		int n = sc.nextInt();
		int q = sc.nextInt();
		String[] name = new String[n];
		int[] time = new int[n];
		int[] timeAns = new int[n];
		int ans = 0;
		Arrays.fill(timeAns, 0);
		for(int i = 0; i < n; i++) {
			name[i] = sc.next();
			time[i] = sc.nextInt();
			Pair pair = new Pair(name[i], time[i]);
			que.add(pair);
		}

		Pair pair;
		while((pair = que.poll()) != null) {
			if(pair.getTime() - q > 0) {
				Pair tmp = new Pair(pair.getName(), pair.getTime() - q);
				que.add(tmp);
				ans += q;
			}else {
				System.out.println(pair.getName() + " " + (pair.getTime() + ans));
				ans += pair.getTime();
			}
		}
	}
	
}