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;
long long T, N, M, a, b, c, d;
long long t[100100];
long long ans[100100];
vector<pair<long long, long long>> p;
long long tr = 1000000;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  long long i;
  for (i = 0; i < N; i++) {
    cin >> t[i];
    ans[i] = tr;
    tr--;
    if (i == 0 && t[i] >= 2) {
      cout << -1;
      return 0;
    }
    if ((!p.size() && t[i]) || (p.size() && p[p.size() - 1].first != t[i])) {
      p.push_back({t[i], i});
      if (i)
        ans[i] = t[i - 1];
      else
        ans[i] = 0;
    } else if (i == 0)
      p.push_back({0, -1});
  }
  for (i = 0; i < p.size() - 1; i++) {
    long long j;
    for (j = 1; j < p[i + 1].first - p[i].first; j++) {
      if (j + p[i].second >= p[i + 1].second) {
        cout << -1;
        return 0;
      }
      ans[j + p[i].second] = j + p[i].first;
    }
  }
  for (i = 0; i < N; i++) cout << ans[i] << ' ';
}
