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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <stack>

using namespace std;
using vi = vector<int>;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ITER(it,v) for(auto it=v.begin();it!=v.end();++it)
#define IN(v) if (scanf("%d", &v)<1);
// #define OUT(v) printf("%d\n", v)
#define OUT(v) cout << v << "\n"

int N;
int B[100];
vi ans;
bool dfs(string& s) {
  // s: 抜いた状況
  // 1なら抜いた
  // 深さ優先で抜いてみる
  // OUT(s);
  if (s.find('0') == string::npos)
    return true;
  int index = 1;
  stack<int> st;
  REP(i, N) {
    if (s[i] == '1')
      continue;
    if (B[i] == index) {
      st.push(i);
    }
    ++index;
  }
  while (!st.empty()) {
    int i = st.top(); st.pop();
    s[i] = '1';
    if (dfs(s)) {
      ans.push_back(B[i]);
      return true;
    }
    s[i] = '0';
  }
  return false;
}
int main() {
  IN(N);

  string s;
  REP(i, N) {
    IN(B[i]);
    s += "0";
  }
  if (dfs(s)) {
    ITER(it, ans) {
      OUT(*it);
    }
  } else
    OUT(-1);

  return 0;
}