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 N, arr[100000], hasil[100000], start_prefix = 0;
bool impossible = false;
void fill(int a, int b, int z) {
  int tmp = a;
  for (int i = start_prefix; i < z; i++) {
    if (hasil[i] == -1) {
      start_prefix = i;
      hasil[i] = tmp;
      tmp++;
      if (tmp == b) {
        return;
      }
    }
  }
  impossible = true;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> arr[i];
  int curr_MEX = 0;
  for (int i = 0; i < N; i++) {
    if (arr[i] > i + 1) {
      impossible = true;
      break;
    } else if (arr[i] == curr_MEX) {
      hasil[i] = -1;
    } else if (arr[i] == curr_MEX + 1) {
      hasil[i] = curr_MEX;
      curr_MEX++;
    } else {
      hasil[i] = curr_MEX;
      fill(curr_MEX + 1, arr[i], i);
      curr_MEX = arr[i];
    }
  }
  if (impossible) {
    cout << -1 << '\n';
  } else {
    for (int i = 0; i < N; i++) {
      if (hasil[i] == -1) {
        cout << curr_MEX + 1 << " ";
      } else {
        cout << hasil[i] << " ";
      }
    }
    cout << '\n';
  }
}
