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.Scanner;

class Main {
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int  num = scanner.nextInt();//要素数
		int  time = scanner.nextInt();//要素数
		int[] array = new int[num];
		String[] pro = new String[num];
		
		for(int i = 0; i < num; i ++ ){
			pro[i] = scanner.next();
			array[i] = scanner.nextInt();
		}
		
		Queue(pro, array,num,time);
		
		scanner.close();
	}

	public static void Queue( String[] pro, int[] array,int num,int time){
		int sub;
		int i;
		int end_count = 0;
		int end_count_now;
		int time_count = 0;
		int judge[] = new int[num];
		boolean judge_end = false;
		int count = 0;
		
		while( end_count < num  ){
			end_count_now = 0;
			judge_end = false;

			//System.out.println(count +  "週目");		//出力
			
			for( i = 0; i < num - end_count; i ++ ){
				sub = array[i] - time;
				if(sub <= 0){
					time_count += array[i];
					System.out.println(pro[i] +  " " + time_count);		//出力
					end_count_now ++;
					judge_end = true;
				}else{
					array[i] -= time;
					time_count += time;
					if(judge_end) {
						array[i - end_count_now] = array[i];
						pro[i - end_count_now] = pro[i];
					}
				}
				continue;
			}
			end_count += end_count_now;
			count ++;
		}
	}
}