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 n;
  cin >> n;
  long long a[n];
  vector<long long> v;
  long long p = 0, c = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (i == 0) {
      if (a[i] == 1) {
        v.push_back(0);
      }
    } else {
      if (a[i] == a[i - 1]) c++;
      if (a[i] - a[i - 1] > 1) {
        long long e = a[i] - a[i - 1] - 1;
        if (c >= e) {
          c -= e;
        } else {
          p = 1;
        }
        for (long long d = a[i - 1] + 1; d < a[i]; d++) {
          v.push_back(d);
        }
      }
    }
  }
  if (p == 1 || a[0] > 1) {
    cout << -1 << endl;
  } else {
    c = 0;
    long long e = v.size();
    for (long long i = 0; i < n; i++) {
      if (i == 0) {
        if (a[i] == 1) {
          cout << v[c] << " ";
          c++;
        } else {
          if (c < e) {
            cout << v[c] << " ";
            c++;
          } else
            cout << n + 1 << " ";
        }
      } else {
        if (a[i] == a[i - 1]) {
          if (c < e) {
            cout << v[c] << " ";
            c++;
          } else
            cout << n + 1 << " ";
        } else {
          cout << a[i - 1] << " ";
        }
      }
    }
    cout << endl;
  }
}
