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;
const long long oo = 1e18;
const double PI = acos(-1);
const long long M = 1e9 + 7;
const int N = 10000010;
long long oor(long long a, long long b) { return (a | b); }
int a[100010], b[100010], use[100010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i <= n; ++i) b[i] = -1;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i && a[i] != a[i - 1])
      b[i] = a[i - 1], use[a[i - 1]] = 1;
    else if (i == 0 && a[i])
      b[i] = 0, use[0] = 1;
  }
  use[a[n - 1]] = 1;
  int co = 0;
  for (int i = 0; i < n; ++i) {
    while (use[co]) co++;
    if (b[i] == -1) b[i] = co, use[co] = 1;
    cout << b[i] << ' ';
  }
}
