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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(var, begin, end) for (int var = (begin); var <= (end); var++)
#define RFOR(var, begin, end) for (int var = (begin); var >= (end); var--)
#define REP(var, length) FOR(var, 0, length - 1)
#define RREP(var, length) RFOR(var, length - 1, 0)
#define EACH(value, var) for (auto value : var)
#define SORT(var) sort(var.begin(), var.end())
#define REVERSE(var) reverse(var.begin(), var.end())
#define RSORT(var) SORT(var); REVERSE(var)

const ll MOD = 1e9 + 7;

void solve(istream& cin, ostream& cout) {
  int n;
  cin >> n;
  vector<int> cb(n + 1, 0);
  vector<int> ans;
  REP(i, n) {
    int b;
    cin >> b;
    cb[b]++;
  }
  FOR(i, 1, n) {
    bool ok = false;
    FOR(j, 1, i) {
      if (cb[j]) {
        ok = true;
        cb[j]--;
        ans.push_back(j);
        break;
      }
    }
    if (!ok) {
      cout << -1 << endl;
      return;
    }
  }
  EACH(v, ans) {
    cout << v << endl;
  }
}

#ifndef TEST
int main() {
  solve(cin, cout);
}
#endif