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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
inline bool equals(long double a, long double b) { return fabs(a - b) < 1e-9; }
inline long long gcd(long long a, long long b) {
  a = abs(a);
  b = abs(b);
  while (a > 0 && b > 0) (a > b ? a %= b : b %= a);
  return (a == 0 ? b : a);
}
inline long long power(long long x, long long n, long long m = LLONG_MAX) {
  long long res = 1;
  x = (x % m + m) % m;
  while (n) {
    if (n & 1) res = (res * x) % m;
    x = (x * x) % m;
    n >>= 1;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  set<long long> s;
  map<long long, long long> m;
  for (long long i = 0; i <= n; i++) {
    s.insert(i);
  }
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    if (s.find(arr[i]) != s.end()) s.erase(arr[i]);
    m[arr[i]] = i;
  }
  long long ans[n];
  for (long long i = 0; i < n; i++) {
    ans[i] = -1;
  }
  for (auto kvp : m) {
    long long k = kvp.first;
    long long v = kvp.second;
    if (v < (n - 1)) ans[v + 1] = k;
  }
  for (long long i = 0; i < n; i++) {
    if (ans[i] == -1) {
      long long d = *(s.begin());
      ans[i] = d;
      s.erase(d);
    }
    cout << ans[i] << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 1; i <= t; ++i) {
    solve();
  };
  return 0;
}
