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 PYTHON):

Solution:

class Queue:
    def __init__(self, n):
        self.A = [0 for i in range(n)]
        self.head = 0
        self.tail = 0
        self.len = n

    def is_empty(self):
        return self.head == self.tail

    def is_full(self):
        return (self.tail + 1) % self.len == self.head

    def enqueue(self, x):
        if not self.is_full():
            self.A[self.tail] = x
            self.tail += 1
            if self.tail == self.len:
                self.tail = 0

    def dequeue(self):
        if not self.is_empty():
            x = self.A[self.head]
            self.head += 1
            if self.head == self.len:
                self.head = 0
            return x


class Task:
    def __init__(self, name, time):
        self.name = name
        self.time = time


if __name__ == '__main__':
    Q = Queue(100005)
    n, q = map(int, raw_input().split())
    for i in range(n):
        l = raw_input().split()
        t = Task(l[0], int(l[1]))
        Q.enqueue(t)

    cnt = 0
    while not Q.is_empty():
        t = Q.dequeue()
        if t.time > q:
            cnt += q
            t.time -= q
            Q.enqueue(t)
        else:
            cnt += t.time
            print t.name, cnt

    pass