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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <ctype.h>

int main(void) {
	int n, b[101], a[101], count;

  if(scanf("%d", &n) == 1);
	count = n;

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

	for(int i = n; i > 0; i--) {
		if(i == b[i]) {
			a[n] = i;
			for(int j = i; j < n; j++) {
				b[j] = b[j + 1];
			}
			n--;
			i = n + 1;
		}
	}

	if(n == 0) {
		for(int i = 1; i <= count; i++) {
			printf("%d\n", a[i]);
		}
	} else {
		printf("-1");
	}

  return 0;
}