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.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.Queue;

public class Main {

	public static void main(String[] args) throws IOException {
		// TODO ?????????????????????????????????????????????
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String[] strArray = br.readLine().split(" ");
		
		int n = Integer.parseInt(strArray[0]);
		int q = Integer.parseInt(strArray[1]);
		
		Queue<Process> queue = new ArrayDeque<Process>();
		for(int i = 0; i < n ;i++){
			strArray = br.readLine().split(" ");
			queue.add(new Process(strArray[0], Integer.parseInt(strArray[1])));
		}
		
		int currentTime = 0;
		while(queue.size() > 0){
			if(queue.element().remainTime > q){
				queue.element().remainTime -= q;
				queue.add(queue.poll());
				currentTime += q;
			}
			else{
				currentTime += queue.element().remainTime;
				System.out.println(queue.element().name + " " + currentTime);
				queue.poll();
			}
			
		}
	}

}

class Process {
	String name;
	int remainTime;
	int finishedTime;
	
	public Process (String name, int time){
		this.name = name;
		this.remainTime = time;
	}
}