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;
const int N = 1e5 + 20;
const int M = 1e6 + 20;
int a[N], ans[N];
int vis[M];
int main() {
  cout.tie(0);
  int n;
  scanf("%d", &n);
  for (int i = 0; i <= 1e6; i++) {
    vis[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    vis[a[i]]++;
  }
  bool flag = true;
  int k = 0, cnt = 0;
  while (vis[k] >= 1) {
    k++;
  }
  if (k < a[0] - 1) {
    cout << -1 << "\n";
    return 0;
  }
  vis[k] = 1;
  ans[cnt++] = k;
  for (int i = 1; i < n; i++) {
    if (a[i] > a[i - 1]) {
      if (a[i - 1] == a[i] - 1) {
        k++;
      }
      if (k < a[i] - 1) {
        flag = false;
        break;
      }
      ans[cnt++] = a[i - 1];
    } else if (a[i] == a[i - 1]) {
      while (vis[k] >= 1) {
        k++;
      }
      vis[k] = 1;
      ans[cnt++] = k;
    } else {
      flag = false;
      break;
    }
  }
  if (flag) {
    for (int i = 0; i < cnt; i++) {
      cout << ans[i] << " ";
    }
  } else {
    cout << "-1";
  }
  return 0;
}
