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(2)
using namespace std;
const int maxn = 1e6 + 10;
int n, a[maxn], b[maxn];
int vis[maxn], now = 0;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    vis[a[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (i >= 2 and vis[a[i - 1]] == 0) {
      b[i] = a[i - 1];
      if (now == b[i]) now++;
    } else {
      while (vis[now]) now++;
      b[i] = now;
      now++;
    }
    vis[a[i]]--;
  }
  bool t = 1;
  memset(vis, 0, sizeof(vis));
  now = 0;
  for (int i = 1; i <= n; i++) {
    vis[b[i]]++;
    while (vis[now]) now++;
    if (now != a[i]) {
      t = 0;
      break;
    }
  }
  if (!t) {
    cout << -1 << '\n';
    return 0;
  }
  for (int i = 1; i <= n; i++) cout << b[i] << ' ';
  cout << '\n';
  return 0;
}
