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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, ans = 0, i, j, k = 1000000, x;
  cin >> n;
  long long int a[n + 1], b[n + 1];
  long long int m[100010];
  for (i = 0; i < 100010; i++) {
    m[i] = 0;
  }
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > i) {
      ans = 1;
    }
  }
  if (ans == 1) {
    cout << -1;
  } else {
    x = a[1];
    if (a[1] == 1) {
      b[1] = 0;
    } else {
      b[1] = -1;
    }
    for (i = 2; i <= n; i++) {
      if (a[i] != a[i - 1]) {
        m[x] = 1;
        b[i] = x;
        x = a[i];
      } else {
        b[i] = -1;
      }
    }
    x = a[n];
    x--;
    for (i = n; i >= 1; i--) {
      while (x >= 0 && m[x] == 1) {
        x--;
      }
      if (x < 0) {
        break;
      }
      if (m[x] == 0 && a[i] < x && b[i] == -1) {
        b[i] = x;
        m[x] = 1;
      }
    }
    for (i = 0; i < a[n]; i++) {
      if (m[i] == 0) {
        ans = 1;
      }
    }
    for (i = 1; i <= n; i++) {
      if (b[i] == -1) {
        cout << k << " ";
      } else {
        cout << b[i] << " ";
      }
    }
  }
  return 0;
}
