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 N = 2e5 + 30, Mod = 1e9 + 7;
const long long SQ = 330;
long long a[N], b[N], f[N], use[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long n;
  cin >> n;
  fill(b, b + N, -1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = n - 1; i; i--) {
    if (a[i] == a[i + 1])
      f[i] = f[i + 1];
    else
      f[i] = a[i + 1];
  }
  if (n == 1) {
    if (a[1] == 0)
      cout << 1;
    else
      cout << 0;
    exit(0);
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] != a[i + 1]) {
      b[i + 1] = a[i];
      use[a[i]] = 1;
    }
  }
  set<long long> s;
  for (int i = 0; i <= n; i++) {
    if (!use[i]) s.insert(i);
  }
  for (int i = 1; i <= n; i++) {
    if (~b[i]) continue;
    b[i] = *s.begin();
    s.erase(s.begin());
  }
  for (int i = 1; i <= n; i++) cout << b[i] << ' ';
}
