Problem:
Score : 400 points

Problem Statement

We have N bricks arranged in a row from left to right.

The i-th brick from the left (1 \leq i \leq N) has an integer a_i written on it.

Among them, you can break at most N-1 bricks of your choice.

Let us say there are K bricks remaining. Snuke will be satisfied if, for each integer i (1 \leq i \leq K), the i-th of those brick from the left has the integer i written on it.

Find the minimum number of bricks you need to break to satisfy Snuke's desire. If his desire is unsatisfiable, print -1 instead.

Constraints

All values in input are integers.

1 \leq N \leq 200000

1 \leq a_i \leq N

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of bricks that need to be broken to satisfy Snuke's desire, or print -1 if his desire is unsatisfiable.

Sample Input 1

3
2 1 2

Sample Output 1

1

If we break the leftmost brick, the remaining bricks have integers 1 and 2 written on them from left to right, in which case Snuke will be satisfied.

Sample Input 2

3
2 2 2

Sample Output 2

-1

In this case, there is no way to break some of the bricks to satisfy Snuke's desire.

Sample Input 3

10
3 1 4 1 5 9 2 6 5 3

Sample Output 3

7

Sample Input 4

1
1

Sample Output 4

0

There may be no need to break the bricks at all.

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

n = gets.to_i
nums = gets.split(' ').map(&:to_i)
count = 0
accurate_num_count = 0
n.times do |a|
    #大丈夫なところは飛ばしたい現状初めからいちいち見てしまう。
    # puts 'a=', a
        accurate_num_count..(nums.length).times do |b| 
        if nums[b] == b+1
            next
        else
            # puts 'b=', b
            nums.delete_at(nums.index(nums[b]))
            count += 1
            accurate_num_count = b
            break
        end
    end
end
if nums[0] != 1
    puts -1
else
    puts count
end