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 MOD = 998244353;
long long binary_exp(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % MOD;
    }
    a = (a * a) % MOD;
    b = b >> 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long T;
  T = 1;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> v(n + 1, 0);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      v[a[i]]++;
    }
    set<int> s;
    for (int i = 0; i <= n; i++) {
      if (v[i] == 0) {
        s.insert(i);
      }
    }
    auto it = s.begin();
    vector<int> b(n);
    b[0] = *it;
    it++;
    for (int i = 1; i < n; i++) {
      if (a[i] != a[i - 1]) {
        b[i] = a[i - 1];
      } else {
        b[i] = *it;
        it++;
      }
    }
    for (int i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
