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 int mod = 1000000007;
long long int a[100004], b[100004], vis[100004], pos[100004], ct[100004];
void solve() {
  long long int n, ma = 0, idx, curr;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    ma = max(ma, a[i]);
    ct[a[i]]++;
  }
  vector<long long int> v;
  for (long long int i = 0; i <= ma; i++) {
    if (ct[i] == 0) {
      v.push_back(i);
    }
  }
  idx = n;
  for (long long int i = n - 1; i >= 0; i--) {
    if (a[i] == curr) continue;
    curr = a[i];
    b[idx] = curr;
    vis[idx] = 1;
    idx = i;
  }
  long long int sz = v.size(), j = 0, ele = 1000000;
  for (long long int i = 0; i < n; i++) {
    if (vis[i] == 0) {
      if (j < sz) {
        b[i] = v[j];
        j++;
      } else
        b[i] = ele;
    }
  }
  if (j != sz) {
    cout << -1 << "\n";
  } else {
    for (long long int i = 0; i < n; i++) cout << b[i] << " ";
    cout << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  ;
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
