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;

class Task{
    String name;
    int time;
}

public class Main {

    static final int len = 100005;
    static Task[] num = new Task[len];

    static int head = 0;
    static int tail = 0;

    // ??????????????????????????????
    public static void enqueue(Task task){
        num[tail] = task;
        tail = (tail+1) % len;
    }

    // ??????????????????????????????
    public static Task dequeue(){
        Task temp = num[head];
        head = (head+1) % len;
        return temp;
    }

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int q = sc.nextInt();
        for (int i=0; i<n; i++){
            Task temp = new Task();
            temp.name = sc.next();
            temp.time = sc.nextInt();
            enqueue(temp);
        }

        tail = n;

        int sum = 0;

        //????????????????????¶????????¬??????????????????
        while (head != tail){
            Task temp = dequeue();
            if (temp.time <= q){
                sum += temp.time;
                System.out.println(temp.name + " " + sum);
            }
            else {
                temp.time -= q;
                enqueue(temp);
                sum += q;
            }
        }
    }

}