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;
using ll = long long;
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
  }
  int k = v[n - 1] + 2;
  vector<int> root(k, 0);
  vector<int> ans(n, -1);
  for (int i = 1; i < n; ++i) {
    root[v[i]] = 1;
    if (v[i] != v[i - 1]) {
      ans[i] = v[i - 1];
    }
  }
  int que = 0;
  for (int i = 0; i < n; ++i) {
    if (ans[i] == -1) {
      while (root[que] == 1 && que <= v[n - 1]) {
        ++que;
      }
      ans[i] = que;
      ++que;
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d ", ans[i]);
  }
}
int main(void) { solve(); }
