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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <queue>
#include <map>
#include <numeric>
#include <unordered_map>
#include <iomanip>
#include <functional>
#include <bitset>
#include <complex>

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define all(x) (x).begin(),(x).end()

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

typedef long long ll;

using namespace std;


int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin>>n;
  vector<int>a(n),da(n),ans(n),used(n);
  rep(i,n)cin>>a[i];
  rep(i,n)da[i]=i+1-a[i];
  //rep(i,n)cout<<da[i]<<" ";
  rep(i,n)
  {
    int max0=-1;
    for(int j=n-1;j>=0;j--)
    {
      if(da[j]==0&&!used[j])
      {
        max0=j;
        used[j]=1;
        break;
      }
    }
    if(max0==-1)
    {
      cout<<-1<<endl;
      return 0;
    }
    ans[i]=a[max0];
    repi(j,max0+1,n)
    {
      da[j]--;
    }
  }
  for(int i=n-1;i>=0;i--)
  {
    cout<<ans[i]<<"\n";
  }
  cout<<flush;
  return 0;

}