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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  vector<int> a(n), b(n + 1, -1);
  set<int> st, stInd;
  for (int i = 0; i < n; i++) {
    st.insert(i);
    stInd.insert(i);
  }
  st.insert(n);
  ;
  for (auto &x : a) cin >> x;
  for (int i = 1; i <= n; i++) {
    if (a[i - 1] != (i == n ? 1e9 : a[i])) {
      b[i] = a[i - 1];
      st.erase(b[i]);
      stInd.erase(i);
    }
  }
  while (!st.empty()) {
    auto val = st.begin();
    auto ind = stInd.begin();
    b[*ind] = *val;
    st.erase(val);
    stInd.erase(ind);
  }
  for (int i = 0; i < n; i++) cout << b[i] << " ";
  cout << '\n';
}
