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: Time Limit Exceeded)

#include <stdio.h>

static int N;
static int B[100];

int search(int i, int *data) {
  if(i == 0) {
    return 1;
  }

  for(int j = (i-1); j >= 0; j--) {
    if(data[j] == (j+1)) {
      int newdata[100];
      int dst = 0;
      for(int k = 0; k < i; k++) {
	if(k != j)
	  newdata[dst++] = data[k];
      }
      int ret = search(i-1, newdata);
      if(ret == 1) {
	printf("%d\n", (j+1));
	return 1;
      }
    }
  }
  return 0;
}

int main(void) {
  scanf("%d", &N);

  int Num[100];
  for(int i = 0; i < N; i++) {
    Num[i] = 0;
  }
  for(int i = 0; i < N; i++) {
    scanf("%d", &B[i]);
    Num[B[i]-1]++;
  }
  int j = 1;
  for(int i = N-1; i > 0; i--, j++) {
    if(Num[i] > j) {
      printf("-1\n");
      return 0;
    }
  }

  if(search(N, B) == 0) {
    printf("-1\n");
  }

  return 0;
}