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.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.Queue;
import java.util.ArrayDeque;
import java.util.Map;
import java.util.LinkedList;

public class Main{
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String[] line = br.readLine().split(" ");
    int n = Integer.parseInt(line[0]);
    int q = Integer.parseInt(line[1]);
    LinkedList<String> queue = new LinkedList<String>();
    int time = 0;
    int cnt = n;
    Map<String, Integer> map = new HashMap<>();

    for(int i = 0; i < n; i++){
      line = br.readLine().split(" ");
      map.put(line[0], Integer.parseInt(line[1]));
      queue.add(line[0]);
    }

    while(true){
      String x = queue.poll();
      if(map.get(x) > q){
        time += q;
        map.put(x, map.get(x) - q);
        queue.add(x);
      }else{
        time += map.get(x);
        map.put(x, time);
        System.out.println(x + " " + map.get(x));
        n--;
      }
      if(n == 0)  break;
    }
  }
}