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 visited[1000000] = {};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  ;
  long long test = 1;
  cin >> test;
  while (test-- > 0) {
    long long n;
    cin >> n;
    vector<long long> v(n + 1), ans(n + 1, -1);
    v[0] = 0;
    long long last_ind = 0;
    for (long long i = 1; i <= n; i++) {
      cin >> v[i];
    }
    visited[v[n]] = 1;
    for (long long i = 1; i <= n; i++) {
      if (v[i] > v[i - 1]) {
        ans[i] = v[i - 1];
        visited[v[i - 1]] = 1;
      }
    }
    long long val = 0;
    for (long long i = 1; i <= n; i++) {
      if (ans[i] == -1) {
        while (visited[val] > 0) {
          val++;
        }
        ans[i] = val;
        visited[val] = 1;
      }
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
}
