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;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
int a[100005], b[100005];
signed main() {
  int n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  vector<int> c;
  for (int i = 1; i <= n; i++) {
    if (a[i] == a[i - 1])
      c.push_back(i);
    else {
      if (a[i - 1] + c.size() + 1 < a[i]) {
        cout << "-1\n";
        return 0;
        ;
      }
      for (int j = 0; j < a[i] - a[i - 1] - 1; j++) {
        b[c.back()] = a[i] - j - 1;
        c.pop_back();
      }
      b[i] = a[i - 1];
    }
  }
  for (int i : c) b[i] = n + 1;
  for (int i = 1; i <= n; i++) printf("%d ", b[i]);
  return 0;
}
