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;
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int min(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int power(long long int x, long long int y,
                    long long int m = 1000000007) {
  if (y < 0) {
    x = power(x, m - 2);
    y = -y;
  }
  long long int ans = 1;
  x = x % m;
  while (y) {
    if (y & 1) ans = (ans * x) % m;
    y = y / 2;
    x = (x * x) % m;
  }
  return ans;
}
long long int factorial(long long int n, long long int m = 1000000007) {
  long long int ans = 1;
  long long int i = 1;
  while (i <= n) {
    ans = (ans * i) % m;
    i++;
  }
  return ans;
}
long long int vis[1000000] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  long long int n;
  cin >> n;
  long long int arr[n];
  long long int flag = 1;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    if (i + 1 < arr[i]) flag = 0;
    vis[arr[i]] = 1;
  }
  if (flag == 0) {
    cout << "-1\n";
    return 0;
  }
  vector<long long int> ans;
  long long int p = 0;
  for (long long int i = 0; i < n; i++) {
    if (i != 0 && arr[i] != arr[i - 1]) {
      ans.push_back(arr[i - 1]);
    } else {
      while (vis[p] == 1) p++;
      ans.push_back(p);
      p++;
    }
  }
  long long int len = ans.size();
  for (long long int i = 0; i < len; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
  return 0;
}
