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.LinkedList;
import java.util.Queue; 
import java.util.*;

public class Main {

	public static void main(String argv[]) {

        Scanner in = new Scanner(System.in);
        int numTask = in.nextInt();
        int maxTime = in.nextInt();
        int curr=0;
        in.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        Queue<String> q = new LinkedList<>();
     // accessing the numbers in the second line
        for (int i = 0; i < numTask; i++){
//        	String str = in.nextLine();
        	q.add(in.nextLine());
        }

        while(q.size() > 0){
        	  String element = q.remove();
        	  String[] oneTask = element.split(" ");
        	  int timeDiff = Integer.parseInt(oneTask[1]) - maxTime;
        	  if(timeDiff<=0)
        	  {
        		  curr = curr + Integer.parseInt(oneTask[1]);
        		  System.out.println(oneTask[0] + ' ' + curr);
        	  }
        	  else
        	  {
        		  curr = curr + maxTime;
        		  oneTask[1] = Integer.toString(timeDiff);
        		  q.add(oneTask[0] + ' ' + oneTask[1]);
        	  }
        }

        in.close();
    }
}