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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

int N;
int b[128];

int mareitaso[128];
int yukati[128];

int main(void) {
	int i;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d", &b[i]) != 1) return 1;
	}
	memcpy(mareitaso, b, sizeof(*b) * N);
	for (i = N; i > 0; i--) {
		int j;
		int ok = 0;
		for (j = i - 1; j >= 0; j--) {
			if (mareitaso[j] == j + 1) {
				yukati[i] = j + 1;
				memmove(mareitaso + j, mareitaso + j + 1, sizeof(*mareitaso) * (N - j /* - 1 */));
				ok = 1;
				break;
			}
		}
		if (!ok) {
			puts("-1");
			return 0;
		}
	}
	for (i = 1; i <= N; i++) {
		printf("%d\n", yukati[i]);
	}
	return 0;
}