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 <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const long long inf = 1e18;
const long long N = 2e5 + 10;
long long a[N], b[N];
void solve(long long tc) {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  a[0] = 0;
  set<long long> s, m;
  for (long long i = 1; i <= n; i++) {
    m.insert(a[i]);
  }
  for (long long i = 1; i <= 1e6; i++) {
    if (m.find(i) == m.end()) s.insert(i);
  }
  long long ans[n + 1];
  for (long long i = 1; i <= n; i++) {
    if (a[i] != a[i - 1])
      ans[i] = a[i - 1];
    else {
      long long x;
      s.erase(x);
      if (s.size() == 0)
        x = 1e6;
      else {
        x = *s.begin();
        s.erase(x);
      }
      ans[i] = x;
    }
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve(t);
  }
  return 0;
}
