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;
void solve(int TestCase) {
  int n;
  cin >> n;
  auto A = vector<int>(n);
  for (auto& x : A) cin >> x;
  if (A[0] > 1) {
    cout << -1 << endl;
    return;
  }
  auto ret = vector<int>(n, -1);
  auto q = queue<int>{};
  if (A[0] == 1)
    ret[0] = 0;
  else
    q.push(0);
  for (auto i = 0; i < n - 1; ++i) {
    if (A[i] != A[i + 1]) {
      ret[i + 1] = A[i];
      for (auto j = A[i] + 1; j < A[i + 1]; ++j) {
        if (q.empty()) {
          cout << -1 << endl;
          return;
        }
        ret[q.front()] = j;
        q.pop();
      }
    } else {
      q.push(i + 1);
    }
  }
  int val = *max_element(A.begin(), A.end()) + 1;
  while (!q.empty()) {
    ret[q.front()] = val;
    q.pop();
  }
  for (auto r : ret) cout << r << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (auto i = 1; i <= t; ++i) {
    solve(i);
  }
}
