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.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO ?????????????????????????????????????????????
		Scanner sc = new Scanner(System.in);
		String line1 = sc.nextLine();
		String[] li1 = line1.split(" ");
		int n = Integer.parseInt(li1[0]);
		int q = Integer.parseInt(li1[1]);

		ArrayList<String> st = new ArrayList<String>();
		ArrayList<Integer> in = new ArrayList<Integer>();

		for (int i = 0; i < n; i++) {
			String kari1 = sc.nextLine();
			String[] kari2 = kari1.split(" ");
			
			st.add(kari2[0]);
			in.add(Integer.parseInt(kari2[1]));
		}
		
		int time = 0;
		
		while (in.size() > 0) {
			if (in.get(0) <= q) {
				time += in.get(0);
				System.out.println(st.get(0) + " " + time);
				in.remove(0);
				st.remove(0);
			} else {
				time += q;
				in.add(in.get(0) - q);
				st.add(st.get(0));
				in.remove(0);
				st.remove(0);

			}
		}

	}

}