Problem:
Score : 400 points

Problem Statement

Snuke has an empty sequence a.

He will perform N operations on this sequence.

In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).

You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq b_i \leq N

Input

Input is given from Standard Input in the following format:

N
b_1 \dots b_N

Output

If there is no sequence of N operations after which a would be equal to b, print -1.
If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.

Sample Input 1

3
1 2 1

Sample Output 1

1
1
2

In this sequence of operations, the sequence a changes as follows:

After the first operation: (1)

After the second operation: (1,1)

After the third operation: (1,2,1)

Sample Input 2

2
2 2

Sample Output 2

-1

2 cannot be inserted at the beginning of the sequence, so this is impossible.

Sample Input 3

9
1 1 1 2 2 1 2 3 2

Sample Output 3

1
2
2
3
1
2
2
1
1

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

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

public class Main {
    static MyReader in = new MyReader();

    public static void main(String[] args) {
        int N = in.i();
        int[] b = in.ii(N);
        int[] ans = new int[N];
        for (int i = N - 1, j; i >= 0; i--) {
            for (j = i; j >= 0; j--) {
                if (j + 1 == b[j]) {
                    ans[i] = b[j];
                    for (int k = j + 1; k <= i; k++) {
                        b[k - 1] = b[k];
                    }
                    break;
                }
            }
            if (j == -1) {
                System.out.println(-1);
                return;
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < N; i++) {
            sb.append(ans[i]).append('\n');
        }
        System.out.print(sb);
    }
}

class MyReader extends BufferedReader {
    char[] cbuf = new char[1024];
    int head = 0;
    int tail = 0;

    MyReader() {
        super(new InputStreamReader(System.in));
    }

    char next() {
        if (head == tail) {
            try {
                tail = super.read(cbuf, 0, cbuf.length);
            } catch (IOException e) {
                e.printStackTrace();
            }
            head = 0;
        }
        return cbuf[head++];
    }

    int i() {
        int k = next() - '0';
        boolean minus = k == -3;
        int n = minus ? 0 : k;
        while (0 <= (k = next() - '0') && k <= 9) n = 10 * n + k;
        return minus ? -n : n;
    }

    int[] ii(final int N) {
        int[] a = new int[N];
        for (int j = 0; j < a.length; j++) a[j] = i();
        return a;
    }

    long l() {
        int k = next() - '0';
        boolean minus = k == -3;
        long n = minus ? 0 : k;
        while (0 <= (k = next() - '0') && k <= 9) n = 10 * n + k;
        return minus ? -n : n;
    }

    char[] s(final int N) {
        char[] s = new char[N];
        for (int i = 0; i < N; i++) {
            s[i] = next();
        }
        next();
        return s;
    }

    public int read(char[] cbuf) {
        int i;
        char c;
        for (i = 0; (c = next()) != ' ' && c != '\n'; i++) cbuf[i] = c;
        return i;
    }

    public String readLine() {
        try {
            return super.readLine();
        } catch (IOException e) {
            return null;
        }
    }
}