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;
class C {
 public:
  int num;
  int numr;
  int deno;
};
void dfs(int s, int *vis, int **catche, vector<int> &vec, int n) {
  vis[s] = 0;
  vec.push_back(s);
  for (int i = 0; i < n; i++) {
    if (vis[i] != 0 && catche[s][i] == 1) {
      dfs(i, vis, catche, vec, n);
    }
  }
  return;
}
void testcase() {
  int n;
  cin >> n;
  unordered_map<int, bool> m, m2, m3;
  int *a = new int[n];
  int *b = new int[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  bool ok = true;
  for (int i = n - 1; i >= 0; i--) {
    int size = i + 1;
    int meresechote = a[i] - 0;
    if (size < meresechote) ok = false;
    m[a[i]] = false;
  }
  if (!ok) {
    cout << -1 << "\n";
    return;
  }
  for (int i = n - 1; i >= 1; i--) {
    if (a[i] != a[i - 1]) {
      b[i] = a[i - 1];
      m3[i] = false;
    }
  }
  int num = 0;
  for (int i = 0; i < n; i++) {
    if (m3.count(i) == 0) {
      while (m.count(num) == 1) {
        num++;
      }
      b[i] = num;
      num++;
    }
    cout << b[i] << " ";
  }
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    testcase();
  }
  return 0;
}
