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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.util.function.*;

class Main {
    static In in = new In();
    static Out out = new Out();
    static long mod = 1000000007;
    static long inf = 0xfffffffffffffffL;

    void solve() {
        int n = in.nextInt();
        int[] a = in.nextIntArray(n);
        int j = 1;
        for (int i = 0; i < a.length; i++) {
            if (a[i] == j) {
                j++;
            }
        }
        if (j == 1) {
            out.println(-1);
        } else {
            out.println(n - j + 1);
        }
    }

    public static void main(String[]$) {
        try {
            new Main().solve();
            out.flush();
        } catch (Throwable t) {
            out.flush();
            t.printStackTrace();
        }
    }
}

class In {
    private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 0x10000);
    private StringTokenizer tokenizer;

    String next() {
        try {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
        } catch (IOException ignored) {
        }
        return tokenizer.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    char[][] nextCharArray(int n, int m) {
        char[][] a = new char[n][m];
        for (int i = 0; i < n; i++) {
            a[i] = next().toCharArray();
        }
        return a;
    }

    int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    int[] nextIntArray(int n, IntUnaryOperator op) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsInt(nextInt());
        }
        return a;
    }

    long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextLong();
        }
        return a;
    }

    long[] nextLongArray(int n, LongUnaryOperator op) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsLong(nextLong());
        }
        return a;
    }
}

class Out {
    private static PrintWriter out = new PrintWriter(System.out);

    void println(Object... a) {
        StringJoiner joiner = new StringJoiner(" ");
        for (Object obj : a) {
            joiner.add(String.valueOf(obj));
        }
        out.println(joiner);
    }

    void flush() {
        out.flush();
    }
}