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 Pair {
    String name;
    int time;
}

public class Main {
    static int LEN;
    static int head;
    static int tail;
    static Pair[] pairs;

    // データの追加
    public static void enqueue(Pair x) {
        pairs[tail] = x;
        tail = (tail + 1) % LEN;
    }

    // データを取り出す
    public static Pair dequeue() {
        Pair x = pairs[head];
        head = (head + 1) % LEN;
        return x;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(); // n個のプロセス
        int q = sc.nextInt(); // 最大実行時間
        LEN = 100005;
        pairs = new Pair[LEN];
        for (int i = 1; i <= n; i++) {
            pairs[i] = new Pair();
            pairs[i].name = sc.next();
            pairs[i].time = sc.nextInt();
        }
        head = 1;
        tail = n + 1;
        int now = 0;
        while (tail != head) {
            Pair u = dequeue();
            if (u.time - q > 0) {
                now += q;
                u.time = u.time - q;
                enqueue(u);
            } else {
                now += u.time;
                System.out.println(u.name + " " + now);
            }
        }
        sc.close();
    }

}