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;
const long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    if (a[0] >= 2) {
      cout << "-1" << endl;
      return 0;
    }
    int b[n];
    b[0] = n + 1;
    for (int i = 1; i < n; i++) {
      if (a[i] == a[i - 1])
        b[i] = n + 1;
      else
        b[i] = a[i - 1];
    }
    int mex = 0;
    vector<int> s;
    unordered_set<int> ss;
    for (int i = 0; i < n; i++) {
      if (mex == a[i]) {
        if (b[i] == n + 1) s.push_back(i);
        continue;
      }
      if (b[i] == n + 1) s.push_back(i);
      ss.insert(b[i]);
      while (ss.find(mex) != ss.end()) mex++;
      while (mex < a[i]) {
        ss.insert(mex);
        if (s.size() == 0) {
          cout << "-1" << endl;
          return 0;
        }
        b[s[s.size() - 1]] = mex;
        while (ss.find(mex) != ss.end()) mex++;
        s.pop_back();
      }
    }
    for (int i = 0; i < n; i++) cout << b[i] << ' ';
  }
  return 0;
}
