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;
double tick() {
  static clock_t oldt, newt = clock();
  double diff = 1.0 * (newt - oldt) / CLOCKS_PER_SEC;
  oldt = newt;
  return diff;
}
long long int fastExpo(long long int a, long long int n) {
  long long int res = 1;
  while (n) {
    if (n % 2) res = res * a;
    n /= 2;
    a = a * a;
  }
  return res;
}
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
int a[100010], b[100010], vis[100010];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(b, -1, sizeof(b));
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; i++) {
    if (a[i] != a[i + 1]) {
      b[i + 1] = a[i];
      vis[a[i]] = 1;
    }
  }
  vis[a[0]] = 1;
  int j = 0;
  for (int i = 0; i < n; i++) {
    if (b[i] == -1) {
      while (vis[j]) j++;
      b[i] = j;
      vis[j] = 1;
    }
  }
  for (int i = 0; i < n; i++) cout << b[i] << " ";
  cout << endl;
  cerr << "execution time is: " << tick() << "\n";
}
