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>

int main()
{
	int i, N, b[101];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d", &(b[i]));
		
	int j, ans[101] = {};
	for (i = N; i >= 1; i--) {
		for (j = i; j >= 1; j--) if (b[j] >= j) break;
		if (b[j] > j) break;
		else {
			ans[i] = j;
			for (; j < i; j++) b[j] = b[j+1];
		}
	}
	if (i >= 1) {
		printf("-1\n");
		fflush(stdout);
		return 0;
	} else for (i = 1; i <= N; i++) printf("%d\n", ans[i]);
	fflush(stdout);
	return 0;
}