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 <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  typename map<first, second>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
int32_t main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> b;
  b.clear();
  b.resize(n);
  long long f = 1;
  for (long long i = n - 1; i >= 0; i--) {
    if (a[i] > (i + 1)) {
      f = 0;
    }
  }
  if (f == 0) {
    cout << -1 << "\n";
  } else {
    set<long long> second;
    for (long long i = 0; i < a[n - 1]; i++) {
      second.insert(-i);
    }
    long long large = 1e6;
    while (second.find(-large) != second.end()) {
      large--;
    }
    for (long long i = (n - 1); i >= 0; i--) {
      if (second.size() == 0) {
        b[i] = large;
        continue;
      }
      long long top = -(*(second.begin()));
      b[i] = top;
      if (top == a[i]) {
        swap(b[i], b[i + 1]);
      }
      second.erase(second.begin());
    }
    for (long long i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}
