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 n, b[200], ans[200];

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