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.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int q = sc.nextInt();
        Deque<String> name = new ArrayDeque<>();
        Deque<Integer> time = new ArrayDeque<>();
        int totalTime = 0;
        for(int i = 0; i < n; i++){
            name.add(sc.next());
            int tmp = sc.nextInt();
            time.add(tmp);
            totalTime += tmp;
        }
        int sumTime = 0;
        Deque<String> finishName = new ArrayDeque<>();
        Deque<Integer> finishTime = new ArrayDeque<>();
        while(sumTime != totalTime){
            int t = time.poll();
            if(t <= q){
                sumTime += t;
                finishTime.add(sumTime);
                finishName.add(name.poll());
            }else{
                sumTime += q;
                time.add(t - q);
                name.add(name.poll());
            }
        }
        for(int i = 0; i < n; i++){
            System.out.println(finishName.poll() + " " + finishTime.poll());
        }
    }
}