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, h, r, l, i, j, ii, x[1000001], y[1000001], z, g, a, b, c, d,
    k;
vector<long long> v;
map<long long, long long> mp;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x[i];
    mp[x[i]]++;
  }
  for (i = 0; i < n; i++) {
    if (v.size() > 0) {
      cout << v[0] << ' ';
      v.clear();
      mp[x[i]]--;
      if (mp[x[i]] == 0 && x[i] < j) v.push_back(x[i]);
    } else {
      while (mp[j] != 0) j++;
      mp[x[i]]--;
      mp[j]++;
      if (mp[x[i]] == 0 && x[i] < j) v.push_back(x[i]);
      cout << j << ' ';
    }
  }
  return 0;
}
