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 main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  std::vector<int> a(n + 2), b(n + 1, -1), c{-1};
  for (int i = 1; i <= n; ++i) std::cin >> a[i];
  a[n + 1] = n + 1;
  for (int i = 0; i <= n; ++i) {
    if (a[i] < a[i + 1]) {
      b[i] = a[i];
      c.push_back(a[i]);
    }
  }
  c.push_back(n + 1);
  for (int i = 0, j = 0; i < int(c.size()) - 1; ++i) {
    for (int v = c[i + 1] - 1; v > c[i]; --v) {
      while (b[j] != -1) ++j;
      b[j++] = v;
    }
  }
  for (int i = 0; i < n; ++i) std::cout << b[i] << " \n"[i == n - 1];
}
