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.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Stack;

class Main{
	public static void main(String args[]) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int q=sc.nextInt();
		String name[]=new String[n];
		
		String queue[]=new String[1000000];
		int head=0, tail=0;
		int totaltime=0;
		int now=0;
		Map<String,Integer> time=new HashMap<String,Integer>();
		
		for(int i=0;i<n;++i) {
			name[i]=sc.next();
			time.put(name[i],sc.nextInt());
			totaltime+=time.get(name[i]);
			queue[tail]=name[i];
			++tail;
		}

		while(now!=totaltime) {
			if(time.get(queue[head])>q) {
				time.put(queue[head], time.get(queue[head])-q);
				queue[tail]=queue[head];
				++head;
				++tail;
				now+=q;
			}
			else {
				now+=time.get(queue[head]);
				System.out.println(queue[head]+" "+now);
				++head;
			}
		}
		

	}
}