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;
const long long INF = 1e18 + 7;
const long long MAX = 1e5;
const long long MOD = 1e9 + 7;
typedef pair<long long, long long> ii;
typedef vector<long long> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;
typedef vector<vii> vvii;
typedef vector<bool> vb;
signed main() {
  long long n;
  cin >> n;
  ;
  vb app(n + 5, false);
  long long f = -1;
  queue<long long> q;
  vi v;
  bool flag = false;
  for (long long i = 0; i < n; i++) {
    long long a;
    cin >> a;
    ;
    app[a] = true;
    v.push_back(a);
    if (a != q.front() || q.size() == 0) q.push(a);
    if (i + 1 < a) flag = true;
  }
  f = 0;
  if (flag) {
    puts("-1");
    return 0;
  } else {
    for (long long i = 0; i < n; i++) {
      if (!app[f + i] && v[i] == q.front()) {
        cout << f + i << " ";
        continue;
      }
      if (!app[f + i]) {
        cout << q.front() << " ";
        q.pop();
        continue;
      }
      i--;
      f++;
    }
    cout << endl;
  }
  return 0;
}
