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 scan_int() {
  int x;
  if (!scanf("%d", &x)) throw 1;
  return x;
}
int main() {
  int N = scan_int();
  vector<int> A(N);
  for (auto& x : A) x = scan_int();
  vector<int> res(N, -1);
  vector<bool> used(N + 10);
  if (A[0] == 1) {
    res[0] = 0;
    used[0] = true;
  }
  for (int i = 1; i < N; ++i)
    if (A[i] != A[i - 1]) {
      res[i] = A[i - 1];
      used[A[i - 1]] = true;
    }
  used[A[N - 1]] = true;
  int p = 0;
  for (int i = 0; i < N; ++i) {
    if (res[i] == -1) {
      while (used[p]) ++p;
      used[p] = true;
      res[i] = p;
    }
  }
  for (int i = 0; i < N; ++i) {
    if (i) printf(" ");
    printf("%d", res[i]);
  }
  puts("");
}
