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 max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int binaryS(int x, int sb[], int m) {
  int l = 0, r = m;
  int mid = (l + r) / 2;
  while (l < r) {
    mid = (l + r) / 2;
    if (sb[mid] < x)
      l = mid + 1;
    else if (sb[mid] > x)
      r = mid - 1;
    else
      return mid;
  }
  return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  int b[n];
  unordered_map<int, int> mp;
  for (__typeof(0) i = 0; i < n; i++) {
    mp[a[i]]++;
  }
  vector<int> tp;
  for (__typeof(0) i = 0; i < *max_element(a, a + n); i++) {
    if (!mp[i]) tp.push_back(i);
  }
  int m = tp.size();
  int j = 0;
  if (tp.empty()) {
    cout << -1 << endl;
  } else {
    b[0] = tp[j];
    j++;
    bool flag = true;
    for (__typeof(0) i = 0; i < n - 1; i++) {
      if (j < m && (tp.front() < a[i])) {
        flag = false;
        break;
      } else if (a[i] == a[i + 1] && (j < m)) {
        b[i + 1] = tp[j];
        j++;
      } else if (a[i] == a[i + 1] && (j >= m)) {
        b[i + 1] = b[i];
      } else {
        b[i + 1] = a[i];
      }
    }
    if (flag) {
      for (int i = 0; i < n; i++) cout << b[i] << ' ';
      cout << endl;
    } else
      cout << -1 << endl;
  }
  return 0;
}
