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;
const int inf = 1e6;
int main() {
  int n;
  cin >> n;
  vector<int> a(n, 0), b(n, inf);
  for (auto& i : a) cin >> i;
  vector<int> cnt(1e5 + 7, 0);
  for (int i = 0; i < n; ++i) {
    auto id = upper_bound(a.begin(), a.end(), a[i]);
    if (id != a.end()) b[id - a.begin()] = a[i], cnt[a[i]] = 1;
  }
  if (a[0] == 1) b[0] = 0, cnt[0] = 1;
  queue<int> que;
  for (int i = 0; i < n; ++i)
    if (b[i] == inf) que.push(i);
  int j = 0;
  while (j < a.back()) {
    if (cnt[j] == 0) {
      b[que.front()] = j;
      que.pop();
      cnt[j] = 1;
    }
    j++;
  }
  for (auto i : b) cout << i << ' ';
  cout << '\n';
  return 0;
}
