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 Rust (1.15.1) (judge verdict: Wrong Answer)

// Template from https://techracho.bpsinc.jp/jhonda/2019_08_05/78537

// input macro for competive-programming from https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        input_inner!{$s, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };

    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, bytes) => {
        read_value!($iter, String).into_bytes()
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    use std::io::Read;
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();

    println!("{}", solve(&s));
}

macro_rules! test {
    ($($exp_input:expr => $output:expr),* $(,)*) => {
        #[test]
        fn solve_test() {
            $(
                let start = std::time::Instant::now();
                let input = solve($exp_input);
                let end = start.elapsed();
                assert_eq!(input, $output, "{}.{:03}", end.as_secs(), end.subsec_millis());
                assert!(end.as_secs() <= 2, "expected_sec: {}sec, actual_sec: {}sec", 2, end.as_secs());
                assert!(end.as_secs() <= 1, "expected_sec: {}sec, actual_sec: {}sec", 1, end.as_secs());
                assert!(end.subsec_millis() <= 10, "expected_ms: {}ms, actual_ms: {}ms", 10, end.subsec_millis());
                assert!(end.subsec_millis() <= 2, "expected_ms: {}ms, actual_ms: {}ms", 2, end.subsec_millis());
            )*
        }
    };
}

test! {
r#"12"# => r#"1"#,
}

fn f(n: usize) -> usize {
    if n < 2 {
        1
    } else {
        n * f (n - 2)
    }
}

fn solve(src: &str) -> String {
    let mut src_iter = src.split_whitespace();
    input! {
        source = src_iter,
        n: usize,
    }
    let mut ans = 0;

    for i in f(n).to_string().chars().rev() {
        if i != '0' {
            break
        }
        ans += 1;
    }

    ans.to_string()
}