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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int main(){
  ll N,count=0,check=0;

cin>>N;

vector<ll> b(N);
vector<ll> tmp;
for(ll i=0;i<N;i++){
cin>>b[i];
}

if(b[0]!=1){
  printf("%d\n",-1);
  return 0;
}

while(1){
for(ll i=b.size()-1;i>-1;i--){
  if(b[i]==i+1){
    tmp.push_back(b[i]);
    b.erase(b.begin()+i);
    break;
  }
  else if(b[0]!= 1){
    printf("%d\n",-1);
    return 0;
  }
}
if(b.size()==1){

  if(b[0]==1){
    break;
  }
  else{
    printf("%d\n",-1);
    return 0;
  }
}
}

tmp.push_back(1);
for(ll i=N-1;i>=0;i--){
  printf("%ld\n",tmp[i]);
}


  return 0;
}