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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
fn main() {
    input! {
        n:usize,
        A:[usize; n]
    };
    let mut i = 1;
    let mut j = 0;
    let mut m: i64 = (n-1) as i64;
    loop {
        if j == n {
            println!("{}", n-1-m as usize);
            break;
        }
        m -= match A[j..].iter().position(|&x| x==i) {
            Some(k) => {
                i += 1;
                j += k+1;
                k as i64
            },
            None => {
                let t = j;
                j = n;
                n as i64 - t as i64
            }
        };
        if m < 0 {
            println!("-1");
            break;
        }
    }
}