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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int N = 1e5;
int last[N + 1], arr[N + 1];
bool used[2 * N + 1];
void solve() {
  int n;
  cin >> n;
  memset(last, -1, sizeof(last));
  memset(used, false, sizeof(used));
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    last[arr[i]] = i;
  }
  if (arr[0] > 1) {
    cout << -1;
    return;
  }
  int ans[n + 1];
  memset(ans, -1, sizeof(ans));
  int prev = 0;
  for (int i = 0; i < n; i++) {
    if (last[arr[i]] == i) {
      ans[i + 1] = arr[i];
      used[arr[i]] = true;
    }
    if (ans[i] != -1) continue;
    while (used[prev] || last[prev] >= i) prev++;
    ans[i] = prev;
    used[prev] = true;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
