Problem:
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Priority Queue

A priority queue is a data structure which maintains a set $S$ of elements, each of with an associated value (key), and supports the following operations:

$insert(S, k)$: insert an element $k$ into the set $S$

$extractMax(S)$: remove and return the element of $S$ with the largest key

Write a program which performs the $insert(S, k)$ and $extractMax(S)$ operations to a priority queue $S$.
The priority queue manages a set of integers, which are also keys for the priority.

Input

Multiple operations to the priority queue $S$ are given. Each operation is given by "insert $k$", "extract" or "end" in a line. Here, $k$ represents an integer element to be inserted to the priority queue.

The input ends with "end" operation.

Output

For each "extract" operation, print the element extracted from the priority queue $S$ in a line.

Constraints

The number of operations $\leq 2,000,000$

$0 \leq k \leq 2,000,000,000$

Sample Input 1

insert 8
insert 2
extract
insert 10
extract
insert 11
extract
extract
end

Sample Output 1

8
10
11
2

Solution in Ruby (judge verdict: Accepted)

def left(i)
    (i+1)*2-1
end

def right(i)
    (i+1)*2
end

def max_heapify(nodes, i)
    l = left(i)
    r = right(i)
    largest = i
    if l < nodes.length && nodes[l] > nodes[i]
        largest = l
    end
    if r < nodes.length && nodes[r] > nodes[largest]
        largest = r
    end
    if largest != i
        nodes[i],nodes[largest] = nodes[largest],nodes[i]
        max_heapify(nodes, largest)
    end
end

def reverse_heapify(nodes)
    i = nodes.length - 1
    while i > 0 do
        j = (i+1)/2-1
        if nodes[i] > nodes[j]
            nodes[i], nodes[j] = nodes[j], nodes[i]
        else
            break
        end
        i = j
    end
end

nodes = []
loop do
    operation=gets.chomp.split(" ")
    if operation[0] == "insert"
        k = operation[1].to_i
        nodes.push(k)
        reverse_heapify(nodes)
    elsif operation[0] == "extract"
        puts nodes[0]
        nodes[0] = -1
        max_heapify(nodes, 0)
    else
        break
    end
end