Problem:
Score : 700 points

Problem Statement

There is a sequence X of length N, where every element is initially 0. Let X_i denote the i-th element of X.

You are given a sequence A of length N. The i-th element of A is A_i. Determine if we can make X equal to A by repeating the operation below. If we can, find the minimum number of operations required.

Choose an integer i such that 1\leq i\leq N-1. Replace the value of X_{i+1} with the value of X_i plus 1.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq A_i \leq 10^9(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

If we can make X equal to A by repeating the operation, print the minimum number of operations required. If we cannot, print -1.

Sample Input 1

4
0
1
1
2

Sample Output 1

3

We can make X equal to A as follows:

Choose i=2. X becomes (0,0,1,0).

Choose i=1. X becomes (0,1,1,0).

Choose i=3. X becomes (0,1,1,2).

Sample Input 2

3
1
2
1

Sample Output 2

-1

Sample Input 3

9
0
1
1
0
1
2
2
1
2

Sample Output 3

8

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class FastPriorityQueue
  attr_reader :size

  def initialize(order: :asc)
    @size = 0
    @nodes = []
  end

  def empty?
    @size.zero?
  end

  def push(node)
    index = size
    @size += 1

    while index > 0
      parent = (index - 1) / 2

      break if @nodes[parent].last <= node.last

      @nodes[index] = @nodes[parent]
      index = parent
    end

    @nodes[index] = node
  end

  alias_method :<<, :push

  def pop
    node = @nodes.first
    root = @nodes[size - 1]
    @size -= 1
    index = 0

    while index * 2 + 1 < @size
      left = 2 * index + 1
      right = left + 1

      if right < size && @nodes[right].last < @nodes[left].last
        left = right
      end

      break if @nodes[left].last >= root.last

      @nodes[index] = @nodes[left]
      index = left
    end

    @nodes[index] = root
    node.first
  end

  def top
    @nodes.first
  end
end

N = gets.to_i
A = N.times.map { gets.to_i }

pque = FastPriorityQueue.new

1.upto(N - 1) do |i|
  if A[i] - A[i - 1] > 1
    puts -1
    exit
  end

  next if A[i] < A[i - 1] + 1

  pque.push([i, -A[i - 1]])
end

ans = 0

until pque.empty?
  idx = pque.pop
  ans += 1

  A[idx] -= A[idx - 1] + 1

  if idx < N - 1 && A[idx + 1] > 0
    pque.push([idx + 1, -A[idx]])
  end

  if idx > 0 && A[idx] > 0 && A[idx] >= A[idx - 1] + 1
    pque.push([idx, -A[idx - 1]])
  end
end

if A.max > 0
  puts -1
else
  puts ans
end