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.*;
import java.util.concurrent.LinkedBlockingQueue;

class Main {
    public static void main (String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
        String[] input = br.readLine().split(" ");
        int n = Integer.parseInt(input[0]);
        int q = Integer.parseInt(input[1]);
        String[] names = new String[n];
        LinkedBlockingQueue<int[]> processes = new LinkedBlockingQueue<int[]>();
        int[] currentProcess;
        int time = 0;
        StringBuilder sb = new StringBuilder();

        for (int i = 0; i < n; i++) {
            input = br.readLine().split(" ");
            names[i] = input[0];
            processes.offer(new int[] {i, Integer.parseInt(input[1])});
        }

        while ((currentProcess = processes.poll()) != null) {
            if (currentProcess[1] <= q) {
                time += currentProcess[1];
                sb.append(names[currentProcess[0]]).append(" ").append(time).append("\n");
            } else {
                time += q;
                currentProcess[1] -= q;
                processes.offer(currentProcess);
            }
        }
        System.out.print(sb.toString());
    }
}