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.*;

class Main {

  static int LENGTH = 100000;
  static int [] queue = new int[LENGTH];
  static String [] name = new String[LENGTH];
  static int tail = 0;
  static int head = 0;
  public static void main(String args[]){
    //宣言
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int max = sc.nextInt();
    int out;
    int count=0;
    //input
    for(int i = 0; i<n; i++){
      name[i] = sc.next();
      queue[i] = sc.nextInt();
      tail++;
    }
    //loop
    while(isEmpty() != true){
      out = dequeue();
      if((out-max)>0){
        enqueue(out-max);//
        count += max;
      }else{
        count += out;
        System.out.println(name[head-1]+" "+count);
      }
    }
  }
  //enqueue
  static void enqueue(int num){
    queue[tail] = num;
    if(head != 0){
    name[tail] = name[head-1];//
  }else{
    name[tail] = name[LENGTH-1];
  }
    if(tail == LENGTH-1){
      tail=0;
    }else{
      tail++;
    }
  }
  //dequeue
  static int dequeue(){
    if(head == LENGTH-1){
      head = 0;
      return queue[LENGTH-1];
    }
    return queue[head++];
  }
  //キューの中身がなくなったかどうかの判定
  static boolean isEmpty(){
    return head == tail;
  }

}