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;
const long long mod = 1000000007;
const long long MAXX = 1e6 + 5;
long long t, n, x;
long long a[MAXX];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (n == 1) {
    if (a[0] == 0) {
      cout << "1" << endl;
      return 0;
    }
    cout << "0" << endl;
    return 0;
  }
  vector<long long> pos(n + 2, 0);
  vector<long long> b(n + 2, 0);
  for (long long i = 0; i <= n; i++) {
    b[i] = i;
    pos[i] = i;
  }
  long long y = 0;
  for (long long i = 1; i < n; i++) {
    if (a[i] != a[i - 1]) {
      swap(b[pos[a[i - 1]]], b[i]);
      swap(pos[a[i - 1]], pos[i]);
      continue;
    }
  }
  long long i = n;
  swap(b[pos[a[i - 1]]], b[i]);
  swap(pos[a[i - 1]], pos[i]);
  bool f = true;
  long long x = 0;
  vector<bool> used(MAXX, false);
  for (long long i = 0; i < n; i++) {
    used[b[i]] = true;
    while (used[x]) {
      x++;
    }
    if (a[i] != x) {
      f = false;
      break;
    }
  }
  if (!f) {
    cout << "-1" << endl;
    return 0;
  }
  for (long long i = 0; i < n; i++) {
    cout << b[i] << " ";
  }
  return 0;
}
