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;
template <typename T>
inline T MIN(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T MAX(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T, typename... Args>
inline T MIN(const T &a, const T &b, Args... args) {
  return MIN(MIN(a, b), args...);
}
template <typename T, typename... Args>
inline T MAX(const T &a, const T &b, Args... args) {
  return MAX(MAX(a, b), args...);
}
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int dir[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<int> res;
  set<int> st;
  for (int i = n; i >= 1; i--) {
    if (a[i] != a[i - 1]) {
      res.push_back(a[i - 1]);
      st.insert(a[i]);
    } else {
      if (i == n) {
        res.push_back(0);
        continue;
      }
      for (int i = res.back() + 1;; i++) {
        if (!st.count(i)) {
          res.push_back(i);
          break;
        }
      }
    }
  }
  reverse(res.begin(), res.end());
  for (auto i : res) cout << i << " ";
  return 0;
}
