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 parseArray(long long* A, long long n) {
  for (long long K = 0; K < n; K++) {
    cin >> A[K];
  }
}
long long modInverse(long long a, long long b) {
  return 1 < a ? b - modInverse(b % a, a) * b / a : 1;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long double dist(long double x, long double y, long double a, long double b) {
  return sqrt((x - a) * (x - a) + (y - b) * (y - b));
}
void debug(long long* a, long long n) {
  for (long long k = 0; k < n; k++) {
    cerr << a[k] << " ";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int n;
  cin >> n;
  long long a[n];
  parseArray(a, n);
  int last_idx[n + 1];
  memset(last_idx, -1, sizeof last_idx);
  for (int k = 0; k < n; k++) last_idx[a[k]] = k;
  vector<long long> ans;
  long long idx = 1;
  if (a[0] == 0)
    ans.push_back(1), idx = 2;
  else
    ans.push_back(0);
  for (int k = 1; k < n; k++) {
    if (a[k] == a[k - 1]) {
      while (idx <= n && last_idx[idx] >= k) idx++;
      ans.push_back(idx++);
    } else {
      ans.push_back(a[k - 1]);
    }
  }
  for (long long e : ans) cout << e << " ";
  cout << endl;
  return 0;
}
