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() {
  int n;
  cin >> n;
  int a[n];
  int flg = 0;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > i + 1) {
      flg = 1;
    }
  }
  if (flg) {
    cout << -1 << endl;
  } else {
    for (int i = 1; i < n; i++) {
      int x = a[i - 1] + 1, y = a[i];
      while (y > x) {
        v.push_back(x);
        x++;
      }
    }
    vector<int> ans;
    int pos = 0, cnt = 0, pos1;
    if (a[pos1] == 0) {
      while (a[pos1] == 0) {
        if (pos == v.size()) {
          ans.push_back(a[n - 1] + 1);
        } else {
          ans.push_back(v[pos]);
          pos++;
        }
        pos1++;
      }
    } else {
      ans.push_back(0);
      pos1++;
      cnt++;
    }
    while (pos1 < n) {
      if (a[pos1] == a[pos1 - 1]) {
        if (pos == v.size()) {
          ans.push_back(a[n - 1] + 1);
        } else {
          ans.push_back(v[pos]);
          pos++;
        }
      } else {
        ans.push_back(cnt);
        if (a[pos1] != a[pos1 - 1] + 1) {
          cnt++;
        }
        cnt++;
      }
      pos1++;
    }
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
