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;
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int main() {
  fastio();
  long long int T = 1;
  for (long long int test = 1; test <= T; test++) {
    long long int x, n, m, k, ans, cnt, sum;
    ans = 0;
    cnt = 0;
    sum = 0;
    cin >> n;
    k = 0;
    x = 0;
    long long int a[n + 1], b[n + 1];
    map<long long int, long long int> mp;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = -1;
    }
    a[n] = 10000000;
    b[n] = -1;
    n++;
    x = 0;
    for (long long int i = 1; i < n; i++) {
      if (a[i] != a[i - 1]) {
        b[i] = a[i - 1];
        mp[a[i - 1]]++;
      }
    }
    x = 0;
    if (a[0] == 1) {
      b[0] = 0;
      mp[0]++;
    }
    for (long long int i = 0; i < n; i++) {
      if (b[i] == -1) {
        while (mp[x] != 0) x++;
        b[i] = x;
        mp[x]++;
      }
      if (i != n - 1) cout << b[i] << " ";
    }
    cout << endl;
  }
}
