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 Common Lisp (SBCL 1.1.14) (judge verdict: Time Limit Exceeded)

#-swank
(unless (member :child-sbcl *features*)
  (quit
   :unix-status
   (process-exit-code
    (run-program *runtime-pathname*
                 `("--control-stack-size" "128MB"
                   "--noinform" "--disable-ldb" "--lose-on-corruption" "--end-runtime-options"
                   "--eval" "(push :child-sbcl *features*)"
                   "--script" ,(namestring *load-pathname*))
                 :output t :error t :input t))))

(defun check-bits (n bits bricks)
      (loop with i = 0
            for j from 0
            for bitidx from (1- n) downto 0
            unless (logbitp bitidx bits)
            do (progn
                 (incf i)
                 (unless (= i (aref bricks j))
                   (return nil)))
            finally (return t)))
     
    (defun make-bit-combinations (n)
      (sort (loop for i from 0 below (1- (expt 2 n))
                  collect i)
            #'<
            :key #'logcount))
     
    (defun solve (n bricks)
      (loop for bits in (make-bit-combinations n)
            if (check-bits n bits bricks)
            do (return (logcount bits))
            finally (return -1)))
     
    #-swank
    (let* ((n (read))
           (bricks (make-array n :element-type 'integer)))
      (loop for i from 0 below n
            do (setf (aref bricks i) (read)))
      (format t "~A~%" (solve n bricks)))