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;
using ll = long long;
using vll = vector<ll>;
using pll = pair<ll, ll>;
using ld = long double;
void solve() {
  ll n;
  cin >> n;
  vll A(n);
  for (int i = 0; i < n; i++) cin >> A[i];
  vll can, vac;
  vll ans(n, -1 * 1LL);
  if (A[0] == 1) ans[0] = 0;
  for (int i = 1; i < n; i++) {
    if (A[i - 1] == A[i])
      vac.push_back(i * 1LL);
    else {
      ans[i] = A[i - 1];
      for (int j = A[i - 1] + 1; j < A[i]; j++) can.push_back(j * 1LL);
    }
  }
  ll j = 0;
  for (int i = 0; i < n && j < can.size(); i++)
    if (ans[i] == -1) ans[i] = can[j++];
  if (j < can.size()) {
    cout << -1 << "\n";
    return;
  }
  for (int i = 0; i < n; i++)
    if (ans[i] == -1 && i == 0)
      ans[i] = ll(1e6);
    else if (ans[i] == -1)
      ans[i] = ans[i - 1];
  for (auto i : ans) cout << i << " ";
}
int main() {
  cout << fixed;
  solve();
}
