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 powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int test_case = 1;
const int maxn = 10100;
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<int> res(n);
  bool doable = true;
  for (long long i = 0; i < n; i++) {
    if (a[i] > i + 1) {
      doable = false;
    }
  }
  if (doable) {
    set<int> s;
    for (long long i = 0; i < n; i++) {
      s.insert(a[i]);
    }
    int proximo = 0;
    while (s.count(proximo)) proximo++;
    res[0] = proximo++;
    for (int i = 1; i < n; i++) {
      if (a[i] > a[i - 1])
        res[i] = a[i - 1];
      else {
        while (s.count(proximo)) proximo++;
        res[i] = proximo++;
      }
    }
    for (long long i = 0; i < n; i++) cout << res[i] << " ";
    cout << "\n";
  } else {
    cout << "-1\n";
  }
  return 0;
}
