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 Python (3.4.3) (judge verdict: Wrong Answer)

n = int(input())
a_list = list(map(int,input().split()))

min_indexlist = [a_list.index(min(a_list))]

m = min(a_list)

for i in range(len(a_list)):
    if( a_list[i] == m+1):
        m = a_list[i]
        min_indexlist.append(i)


#print(min_indexlist)

#砕いたレンガの後のリスト
crash_list = []
for i in range(len(min_indexlist)):
    crash_list.append(a_list[min_indexlist[i]])

#print(crash_list)

#砕いたレンガの数
crash = n - len(min_indexlist)


frag=True

for i in range(len(crash_list)):
    if(crash_list[i] != i+1):
        frag = False


if(frag == True):
    if(crash <= n-1):
        print(crash)
elif(frag == False):
    print(-1)