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 t, n, x, a[600000];
vector<int> pos;
int ans[600000];
bool flag[600000];
int main() {
  scanf("%d", &n);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  if (a[1] == 0) {
    pos.push_back(1);
  } else {
    ans[1] = 0;
    flag[0] = true;
  }
  for (int i = 2; i <= n; i++) {
    if (a[i] == a[i - 1]) {
      pos.push_back(i);
    } else {
      ans[i] = a[i - 1];
      flag[a[i - 1]] = true;
      for (int j = a[i - 1]; j < a[i]; j++) {
        if (!flag[j]) {
          ans[pos[cnt]] = j;
          cnt++;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0 && ans[i] == 0) {
      printf("%d ", 1);
    } else {
      printf("%d ", ans[i]);
    }
  }
}
