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;
int main() {
  long long i, j, k, l, m, n, mx = 0ll;
  cin >> n;
  long long ara[n + 1];
  vector<long long> vec;
  long long kara_ache[n + 1], kader_nilam[n + 1];
  memset(kara_ache, 0, sizeof(kara_ache));
  memset(kader_nilam, 0, sizeof(kader_nilam));
  for (i = 1; i <= n; i++) {
    cin >> ara[i];
    mx = max(mx, ara[i]);
    kara_ache[ara[i]] = 1ll;
  }
  k = mx + 1, l = 0ll;
  long long cur_mex = 0ll, cnt = 0ll;
  for (i = 1; i <= n; i++) {
    while (kader_nilam[cur_mex] == 1ll) cur_mex++;
    if (cur_mex == ara[i]) {
      cnt = 0ll;
      for (j = l; j < mx; j++) {
        if (kara_ache[j] == 0 && kader_nilam[j] == 0 && j != cur_mex) {
          vec.push_back(j);
          kader_nilam[j] = 1ll;
          cnt = 1ll;
          l = j;
          break;
        }
      }
      if (cnt == 0ll) {
        vec.push_back(k);
        k++;
      }
    } else {
      vec.push_back(cur_mex);
      kader_nilam[cur_mex] = 1ll;
      cur_mex++;
    }
  }
  for (i = 0; i < vec.size(); i++) cout << vec[i] << " ";
  cout << endl;
  return 0;
}
