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;
clock_t time_p = clock();
void shahil_005() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
int power(long long x, long long y) {
  if (y == 0) return 1;
  long long p = power(x, y / 2);
  p = (p * p);
  return (y % 2 == 0) ? p : (x * p);
}
int powm(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) {
    return 0;
  }
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int lcm(long long a, long long b) {
  long long lar = max(a, b);
  long long small = min(a, b);
  for (long long i = lar;; i += lar) {
    if (i % small == 0) return i;
  }
}
int gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int bsearch(long long arr[], long long l, long long r, long long x) {
  while (l <= r) {
    long long m = l + (r - l) / 2;
    if (arr[m] == x) {
      return m;
    }
    if (arr[m] < x) {
      l = m + 1;
    } else
      r = m - 1;
  }
  return -1;
}
bool prime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
bool sortbydec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return a.first > b.first;
}
const int N = 1e5 + 5;
const long long MOD = (long long)(1e9 + 7);
const long long INIT = (long long)(1e6 + 1);
void solve() {
  long long n;
  cin >> n;
  long long a[n + 1];
  set<long long> s;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  long long x = 0;
  for (long long i = 1; i <= n; i++) {
    while (1) {
      if (s.find(x) == s.end()) {
        break;
      }
      x++;
    }
    cout << x << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
