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)


(defun maximal-natural-sequence-0 (arr)
  (labels ((rec (i)
		(when (= (1+ i) (length arr))
		  (return-from rec 1))
		(1+ (loop for j from (1+ i) below (length arr)
			  when (= (aref arr j) (1+ (aref arr i)))
			  maximizing (rec j)))))
	  (let ((a (loop for i from 0 below (length arr)
			 when (= (aref arr i) 1) maximizing
			 (rec i))))
	    (if (= a 0)
		-1
		(- (length arr) a)))))

(defun maximal-natural-sequence-1 (arr)
  (let ((table (make-array (length arr))))
    (setf (aref table (1- (length arr))) 1)
    (loop for i from (- (length arr) 2) downto 0 do
	  (setf (aref table i)
		(1+ (loop for j from (1+ i) below (length arr)
			  when (= (aref arr j) (1+ (aref arr i)))
			  maximizing (aref table j)))))
    (loop for i from 0 below (length arr)
	  when (= 1 (aref arr i))
	  maximizing (aref table i))))

(let* ((n (read))
       (a (make-array n))
       (has-1 nil))
  (dotimes (i n)
    (let ((r (read)))
      (when (= r 1) (setf has-1 t))
      (setf (aref a i) r)))
  (unless has-1
    (princ -1)
    (exit))
  (princ (- n (maximal-natural-sequence-1 a))))