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 C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

int n;
int b[105];
int order[105];

int main()
{

	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &b[i]);
	}

	for (int i = n; i >= 1; i--) {
		int j = n;

		while (j > 0 && b[j] != j) {
			j--;
		}

		if (j <= 0) {
			printf("-1\n");
			return 0;
		}

		order[i] = j;

		//???
		while (j <= n) {
			b[j] = b[j + 1];
			j++;
		}

	}

	for (int i = 1; i <= n; i++) {
		printf("%d\n", order[i]);
	}

	return 0;
}