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[100005], nxt[100005], ans[100005];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  nxt[n] = a[n];
  for (int i = n - 1; i >= 1; i--) {
    nxt[i] = nxt[i + 1];
    if (a[i] != a[i + 1]) nxt[i] = a[i + 1];
  }
  if (a[1] == 0)
    cout << 1 << ' ';
  else
    cout << 0 << ' ';
  for (int i = 2; i <= n; i++) {
    int j, x = nxt[i];
    for (j = i; j <= n and a[j] == a[i]; j++) {
      x = max(0, x - 1);
      if (x == a[i]) {
        if (x == 0)
          x = 1;
        else
          x = x - 1;
      }
      cout << x << ' ';
    }
    if (j <= n) cout << a[i] << ' ';
    i = j;
  }
}
