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;
long long int arr[201];
string s;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  int arr[100005];
  for (int i = 0; i < n; i++) cin >> arr[i];
  bool vis[100005];
  memset(vis, false, sizeof(vis));
  vector<int> ans;
  if (arr[0] > 1) {
    cout << -1;
    return 0;
  }
  int prev = ((arr[0] == 0) ? 1 : 0);
  vis[prev] = true;
  ans.push_back(prev);
  for (int i = 1; i < n; i++) {
    if (arr[i] < arr[i - 1]) {
      cout << -1;
      return 0;
    }
    if (arr[i] == arr[i - 1]) {
      ans.push_back(arr[i] + 1);
      prev = arr[i] + 1;
      vis[arr[i] + 1] = true;
    } else {
      if (arr[i] == prev) {
        cout << -1;
        return 0;
      }
      if (arr[i] == arr[i - 1] + 1) {
        ans.push_back(arr[i - 1]);
        vis[arr[i - 1]] = true;
        prev = arr[i - 1];
      } else if (arr[i] == arr[i - 1] + 2) {
        if (!vis[arr[i] - 1] && !vis[arr[i] - 2]) {
          cout << -1;
          return 0;
        }
        if (!vis[arr[i] - 2]) {
          vis[arr[i] - 2] = true;
          prev = arr[i] - 2;
        } else if (!vis[arr[i] - 1]) {
          vis[arr[i] - 1] = true;
          prev = arr[i] - 1;
        }
        ans.push_back(prev);
      } else if (arr[i] - arr[i - 1] > 2) {
        cout << -1;
        return 0;
      }
    }
  }
  for (auto& x : ans) cout << x << " ";
  return 0;
}
