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:
import sys

class Queue:
    class Cell:
        def __init__(self, x, y = None):
            self.data = x
            self.next = y

    def __init__(self):
        self.size = 0
        self.rear = None

    def enqueue(self, x):
        if self.size == 0:
            self.rear = Queue.Cell(x)
            self.rear.next = self.rear
        else:
            new_cell = Queue.Cell(x, self.rear.next)
            self.rear.next = new_cell
            self.rear = new_cell
        self.size += 1

    def dequeue(self):
        if self.size == 0: raise IndexError
        front = self.rear.next
        self.rear.next = front.next
        self.size -= 1
        if self.size == 0: self.rear = None
        return front.data

    def isEmpty(self):
        return self.size == 0

def generator(n1, n2, step=1):
    n = n1
    while n<n2:
        yield n
        n+= step

n, q = map(int, raw_input().split())
 
A = Queue()
for i in generator(0,n):
    name, time = raw_input().split()
    A.enqueue([name, int(time)])

stime = 0
while not A.isEmpty():
    job = A.dequeue()
    if q>=job[1]:
        stime += job[1]
        print job[0], stime
    else:
        stime += q
        job[1] -= q
        A.enqueue(job)