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 int INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
int t[110000];
bool used[110000];
int main() {
  int n, m;
  int i, j, k;
  int a, b, c;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> t[i];
  }
  for (i = 0; i < n; i++) {
    used[t[i]] = true;
    if (t[i] > i + 1) {
      cout << -1 << endl;
    }
  }
  for (i = n - 1; i > 0; i--) {
    t[i] = t[i - 1];
  }
  for (i = 0; i < n - 1; i++) {
    if (t[i] == t[i + 1]) t[i] = -1;
  }
  for (i = 0, j = 0; i < n; i++) {
    if (t[i] != -1) continue;
    while (used[j]) j++;
    t[i] = j;
    used[j] = true;
  }
  array_show(t, n);
}
