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 v[100001];
int poz[1000010];
int rez[100001];
int main() {
  int n, next;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    rez[i] = -1;
  }
  if (v[1] == 1) {
    poz[0] = 1;
    next = 1;
  } else {
    poz[1] = 1;
    next = 2;
  }
  for (int i = 2; i <= n; i++) {
    if (v[i] == v[i - 1]) {
      if (next == v[i]) next++;
      poz[next++] = i;
    } else {
      if (next == v[i]) next++;
      poz[next++] = poz[v[i]];
      poz[v[i]] = 0;
      poz[v[i - 1]] = i;
    }
  }
  for (int i = 0; i < 1000001; i++) {
    if (poz[i] != 0) rez[poz[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (rez[i] == -1) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) cout << rez[i] << ' ';
  return 0;
}
