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;
long long ncrp(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = min(i, r); j >= 1; j--) C[j] = (C[j] + C[j - 1]) % p;
  return C[r];
}
long long mod_exp(long long x, long long y, long long mm) {
  x %= mm;
  if (y == 0)
    return (1);
  else if (y % 2 == 0)
    return (mod_exp((x * x) % mm, y / 2, mm));
  else
    return ((x * mod_exp((x * x) % mm, (y - 1) / 2, mm)) % mm);
}
long long node(long long i, long long j, long long m) { return m * i + j; }
pair<long long, long long> loc(long long z, long long m) {
  return {z / m, z % m};
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, r = 0, e;
  cin >> n;
  vector<long long> z;
  long long a[n], ans[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    ans[i] = 1000000;
  }
  for (long long i = 0; i < n; i++) {
    if (r == a[i])
      z.push_back(i);
    else if (r == a[i] - 1) {
      ans[i] = r;
      r++;
    } else {
      z.push_back(i);
      e = z.size();
      for (long long j = e - 1; j >= 0; j--) {
        ans[z[z.size() - 1]] = r;
        r++;
        z.pop_back();
        if (r == a[i]) break;
        if (z.empty()) break;
      }
      if (r != a[i]) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}
