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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

# n, m, l = map(int, input().split())

# list_n = list(map(int, input().split()))

# n = input()
# list = [input() for i in range(N)

# list = [[i for i in range(N)] for _ in range(M)]

import sys
input = sys.stdin.readline

N = int(input())
A = list(map(int, input().split()))

start = 0
n = 0
for i in range(N):
    if i + 1 in A[start:]:
        start = len(A[:start]) + A[start:].index(i+1) + 1
        n = i + 1
    else:
        break
if n != 0:
    print(N-n)
else:
    print(-1)