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 A[100010];
bool cnt[100010];
int main() {
  srand(int(time(NULL)));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]), cnt[A[i]] = true;
  int mx = 0, mex = 0;
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (mex == A[i]) {
      ++mx;
      while (cnt[mx]) ++mx;
      ans.push_back(mx);
    } else {
      ans.push_back(mex);
      mex = A[i];
      if (mx < A[i] - 1) {
        for (int i = 1; i <= n; i++) cout << A[i] << " ";
        cout << endl;
        cout << -1;
        return 0;
      }
      mx = max(mx, A[i]);
    }
  }
  for (int x : ans) cout << x << " ";
}
