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>

int main(){
	int N, i, j, k;
	scanf("%d", &N);
	int *a = (int *)malloc(sizeof(int) * N);
	int *b = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%d", &b[i]);
		b[i]--;
	}
	for(i = 0; i < N; i++){
		if(b[i] > i){
			printf("-1\n");
			return 0;
		}
	}
	int *ans = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		ans[i] = -1;
	}
	for(i = N - 1; i >= 0; i--){
		for(j = 0, k = 0; ; j++){
			if(ans[j] == -1){
				k++;
			}
			if(k > b[i]){
				ans[j] = b[i];
				break;
			}
		}
	}
	for(i = 0; i < N; i++){
		printf("%d\n", ans[i] + 1);
	}
	return 0;
}