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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
template <class T>
void ckmin(T& a, T b) {
  a = min(a, b);
}
template <class T>
void ckmax(T& a, T b) {
  a = max(a, b);
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> ans(n, -1);
  vector<int> used(2 * n, 0);
  int curr = a[0];
  for (int i = 1; i < n; i++) {
    if (a[i] == curr) continue;
    ans[i] = curr;
    used[curr]++;
    curr = a[i];
  }
  used[curr]++;
  int inc = 0;
  for (int i = 0; i < n; i++) {
    if (ans[i] == -1) {
      while (used[inc]) inc++;
      if (!used[inc]) ans[i] = inc++;
    } else {
      debug() << " ["
              << "a[i]"
                 ": "
              << (a[i]) << "] " << inc;
      if (a[i] > inc) {
        cout << "-1\n";
        return 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
