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;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] tmp =  br.readLine().split(" ");

		int n = Integer.parseInt(tmp[0]);
		int qTime = Integer.parseInt(tmp[1]);

		Process[] pAry = new Process[n+1];

		for(int i = 0 ; i < n ; i++){
			tmp =  br.readLine().split(" ");
			pAry[i] = new Process(tmp[0],Integer.parseInt(tmp[1]));
		}

		boolean isFinished = false;
		int top = 0;
		int tail = n;
		int cureentTime = 0;

		while(!isFinished){

			if(pAry[top].time - qTime <= 0){
				cureentTime += pAry[top].time;
				System.out.println(pAry[top].name + " " + cureentTime );
			}else{
				pAry[top].time -= qTime;
				pAry[tail] = pAry[top];
				tail = (tail + 1) % (n + 1);
				cureentTime += qTime;
			}
			top = (top + 1) % (n + 1);
			if(top == tail){
				isFinished = true;
			}
		}
	}
}

class Process {

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

	public String name;
	public int time;
}