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;
long long powa(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) {
      ans = (ans % 1000000007 * x % 1000000007) % 1000000007;
    }
    x = (x % 1000000007 * x % 1000000007) % 1000000007;
    y = y >> 1;
  }
  return ans;
}
void solve() {
  int n, nei = 0, ni = 0, mx = INT_MIN;
  cin >> n;
  vector<int> a(n), b(n, -1);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  int c = 0;
  for (int i = 0; i < n; ++i) {
    if (nei > a[i]) {
      c = 1;
      break;
    }
    while (nei != a[i]) {
      int it = lower_bound(a.begin(), a.end(), nei) - a.begin();
      if (a[it] == nei) {
        int it1 = upper_bound(a.begin(), a.end(), nei) - a.begin();
        b[it1] = nei;
        nei++;
      } else {
        while (ni < n && b[ni] != -1) {
          ni++;
        }
        b[ni] = nei;
        nei++;
      }
    }
  }
  if (c == 1) {
    cout << -1;
  } else
    for (int i = 0; i < n; ++i) {
      if (b[i] == -1) {
        b[i] = mx + 1;
        mx++;
      }
      cout << b[i] << " ";
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
