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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String str = br.readLine();
		String[] head = str.split(" ", 2);
		int n = Integer.parseInt(head[0]); // ????????????????????°
		int q = Integer.parseInt(head[1]); // ???????????????
		str = br.readLine();
		List<String> name = new ArrayList<>();
		List<Integer> time = new ArrayList<>();
		int top = 0;
		int end = 0;
		int clock = 0;
		for (int i = 0; i < n; i++) {
			String[] data = str.split(" ", 2);
			name.add(data[0]);
			time.add(Integer.parseInt(data[1]));
			end++;
			str = br.readLine();
		}
		while (top == 0 || top != end) {
			if (time.get(top) > q) {
				time.add(time.get(top) - q);
				name.add(name.get(top));
				top++;
				end++;
				clock = clock + q;
			} else {
				clock = clock + time.get(top);
				System.out.println(name.get(top) + " " + clock);
				top++;
			}
		}
	}
}