Given an array a of length n, find another array, b, of length n such that:

  * for each i (1 ≤ i ≤ n) MEX(\\{b_1, b_2, …, b_i\})=a_i. 



The MEX of a set of integers is the smallest non-negative integer that doesn't belong to this set.

If such array doesn't exist, determine this.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i) — the elements of the array a. It's guaranteed that a_i ≤ a_{i+1} for 1≤ i < n.

Output

If there's no such array, print a single line containing -1.

Otherwise, print a single line containing n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ 10^6)

If there are multiple answers, print any.

Examples

Input


3
1 2 3


Output


0 1 2 

Input


4
0 0 0 2


Output


1 3 4 0 

Input


3
1 1 3


Output


0 2 1 

Note

In the second test case, other answers like [1,1,1,0], for example, are valid.
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> in(n);
  for (auto& it : in) cin >> it;
  set<int> unk;
  int cur = 0;
  vector<int> res(n, -1);
  for (int i = 0; i < n; i++) {
    unk.insert(i);
    while (cur < in[i]) {
      if (unk.size() == 0) {
        cout << -1 << endl;
        return;
      }
      res[*unk.rbegin()] = cur;
      cur++;
      unk.erase(*unk.rbegin());
    }
  }
  set<int> all;
  for (int i = 0; i <= n; i++) {
    all.insert(i);
  }
  for (int j = 0; j < n; j++) {
    int i = res[j];
    all.erase(i);
    if (all.empty()) continue;
    if (*all.begin() != in[j]) {
      cout << -1;
      return;
    }
  }
  for (int i : res) {
    if (i == -1) i = n + 1;
    cout << i << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
