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 t, n, a[100010], b[100010];
deque<int> x;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] > i) {
      cout << "-1";
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    if (a[i] == a[i - 1])
      continue;
    else {
      for (int v = a[i - 1] + 1; v < a[i]; v++) x.push_back(v);
    }
  }
  for (int i = 1; i <= n; i++) x.push_back(a[n] + 1);
  for (int i = 1; i <= n; i++) {
    if (a[i] == a[i - 1]) {
      b[i] = x.front();
      x.pop_front();
    } else {
      b[i] = a[i - 1];
    }
  }
  for (int i = 1; i <= n; i++) cout << b[i] << " ";
}
