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;
int n, a[100004], exist[1000004];
void solve() {
  cin >> n;
  memset(exist, 0, sizeof(exist));
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    exist[a[i]] = 1;
  }
  a[0] = 0;
  int cur, le;
  if (a[1] == 0)
    cur = 1, le = 0;
  else
    cur = 0, le = 1;
  vector<int> ans;
  ans.push_back(cur);
  cur = 2;
  for (int i = 2; i <= n; i++) {
    while (exist[cur]) cur++;
    if (a[i] != a[i - 1]) {
      ans.push_back(a[i - 1]);
      le++;
      while (le <= cur && !exist[le]) le++;
      if (a[i] != le) {
        cout << -1;
        return;
      }
    } else {
      ans.push_back(cur++);
    }
  }
  for (auto k : ans) cout << k << ' ';
}
int main() {
  int T;
  T = 1;
  while (T--) solve();
  return 0;
}
