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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#include <math.h>
#include <iomanip>
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF=1001001001;
vector<pair<int64_t,int64_t>>prime_factorize(int64_t x){
  vector<pair<int64_t,int64_t>>p;
  for(int64_t i=2;i*i<=x;i++){
    int cnt=0;
    if(x%i==0){
    while(x%i==0){cnt++;x/=i;}
    p.push_back(make_pair(i,cnt));
    }
  }
  if(x!=1){p.push_back(make_pair(x,1));}
  return p; 
}
int main() {
  int N;
  cin>>N;
  vector<int>b(N);
  for(int i=0;i<N;i++){
    cin>>b[i];
  }
  int n=N;
  vector<int>ans;
  for(int i=0;i<N;i++){
    for(int j=(int)b.size()-1;j>=0;j--){
      if(b[j]==j+1){
        ans.push_back(b[j]);
        b.erase(b.begin()+j); 
        break;
      }
    }
  }
  reverse(ans.begin(),ans.end());
  if(b.size()==0){
    for(int i=0;i<N;i++){
      cout<<ans[i]<<endl;
    }
  } 
  else{cout<<-1<<endl;}
  return 0;
}