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>
using namespace std;

#define int long long
#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()
#define pb push_back
#define mp make_pair
#define mt make_tuple

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;

const int inf = 1LL<<60;
const int mod = 1e9 + 7;
const double eps = 1e-9;

/*{
  最右のi=a[i]を消去
}*/

signed main()
{
  int n;
  cin >> n;

  vi a(n);
  rep(i, n){
    cin >> a[i];
    a[i]--;
  }

  vi ans;
  while(1){
    bool flag = false;
    for(int i = a.size()-1; i >= 0; i--){
      if(i == a[i]){
        ans.pb(i+1);
        a.erase(a.begin()+i);
        flag = true;
        break;
      }
    }
    if(!flag) break;
  }

  if(ans.size() != n){
    cout << -1 << endl;
  }else{
    reverse(all(ans));
    rep(i, n) cout << ans[i] << endl;
  }

  return 0;
}