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.ArrayList;
import java.util.Scanner;


class Process{
	
	private String name;
	private int time;
	public Process(String name,int time){
		this.name = name;
		this.time = time;
	}
	
	public String getName() {
		return name;
	}
	public void setName(String name) {
		this.name = name;
	}
	public int getTime() {
		return time;
	}
	public void setTime(int time) {
		this.time = time;
	}
	
	
}

public class Main {
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int p = sc.nextInt();
		
		ArrayList<Process> ar = new ArrayList<Process>();
		ArrayList<Process> result = new ArrayList<Process>();
		
		for (int i = 0; i < n; i++) {
			ar.add(new Process(sc.next(),sc.nextInt()));
		}
		
		int totalTime=0,i=0;
		Process process;
		while(true){
			if(i<n){
				if((process=ar.get(i)).getTime()<=p){
					totalTime += process.getTime();
					result.add(new Process(process.getName(), totalTime));
					ar.remove(i);
					n--;
				}
				else{
					totalTime += p;
					int temp =ar.get(i).getTime()-p;
					ar.get(i).setTime(temp);
					i++;
				}
			}
			else{
				i=0;
			}
			if(ar.size()==0){
				break;
			}
		}
		
		for(Process rProcess:result){
			System.out.println(rProcess.getName()+ " "+rProcess.getTime());
		}
		
	}
	
}