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;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto &i : a) cin >> i;
  vector<int> f(n + 1, -1);
  for (int i = 0; i < n; ++i) f[a[i]] = i;
  vector<int> who(n + 1, -1);
  for (int i = 0; i <= n; ++i)
    if (f[i] != -1) who[f[i] + 1] = i;
  vector<int> v;
  for (int i = 0; i <= n; ++i)
    if (f[i] == -1) v.push_back(i);
  for (int i = n + 1; i <= 2 * n; ++i) v.push_back(i);
  reverse(v.begin(), v.end());
  vector<int> b(n);
  for (int i = 0; i < n; ++i) {
    if (who[i] == -1) {
      b[i] = v.back();
      v.pop_back();
      continue;
    }
    b[i] = who[i];
  }
  vector<int> mp(4 * n);
  int pt = 0;
  for (int i = 0; i < n; ++i) {
    ++mp[b[i]];
    while (mp[pt]) ++pt;
    if (pt != a[i]) {
      cout << "-1\n";
      break;
    }
  }
  for (auto &i : b) cout << i << " \n"[&i == &b.back()];
  return 0;
}
