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.Arrays;
import java.util.Deque;
import java.util.ArrayDeque;
import java.io.BufferedReader;
import java.io.InputStreamReader;


class Main{
	public static void main(String[] args)throws Exception{
		Scanner scanner = new Scanner(System.in);
		Deque<String> name_stack = new ArrayDeque<>();
		Deque<Integer> time_stack = new ArrayDeque<>();
		int n = scanner.nextInt();
		int q = scanner.nextInt();
		String nothing = scanner.nextLine();
		int all_time = 0;

		for(int i = 0; i < n-1; i++){
			name_stack.offer(scanner.next());
			time_stack.offer(scanner.nextInt());
			nothing = scanner.nextLine();
		}
		
		name_stack.offer(scanner.next());
		time_stack.offer(scanner.nextInt());

		while(!(name_stack.isEmpty())){
			String name = name_stack.poll();
			int time = time_stack.poll();

			if(time <= q){
				all_time += time;
				System.out.println(name + " " + all_time);
			}else{
				all_time += q;
				time -= q;
				name_stack.offer(name);
				time_stack.offer(time);
			}
		}

	}
}