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 n, a[100005], b[100005], poss[100005];
void solve() {
  cin >> n;
  cin >> a[0];
  memset(b, -1, sizeof(b));
  bool possible = true;
  if (a[0] == 1) {
    b[0] = 0;
  }
  long long j = 0;
  for (long long i = 1; i < n; i++) {
    cin >> a[i];
    if (possible) {
      if (a[i] != a[i - 1]) {
        b[i] = a[i - 1];
        for (long long k = a[i - 1] + 1; k < a[i]; k++) {
          while (b[j] != -1) j++;
          if (j >= i) {
            possible = false;
            break;
          }
          b[j] = k;
        }
      }
    }
  }
  for (long long i = j; i < n; i++)
    if (b[i] == -1) b[i] = 1000000;
  if (possible) {
    cout << "" << b[0];
    for (long long noname = 0 + 1; noname < n; noname++)
      cout << " " << b[noname];
    cout << "\n";
    return;
  }
  cout << -1 << endl;
}
void aux() {}
int main() {
  aux();
  long long _ = 1;
  while (_--) {
    solve();
  }
  return 0;
}
