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 printAll(int n, ...) {
  if (1) {
    va_list ap;
    va_start(ap, n);
    for (int i = 1; i <= n; i++) cout << va_arg(ap, long long int) << " ";
    cout << endl;
    va_end(ap);
  }
}
template <typename T>
string toString(T n) {
  stringstream ss;
  ss << n;
  string s;
  ss >> s;
  return s;
}
void init() {
  cout << fixed << setprecision(0);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
bool isInteger(double n) {
  if (n - (long long int)n == 0)
    return true;
  else
    return false;
}
int main() {
  init();
  long long int i, j, k;
  long long int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> ara(n);
    bool skiped = false;
    bool done = false;
    int start, finish;
    start = finish = n + 1;
    for (int i = 0; i < n; i++) {
      cin >> ara[i];
      if (!skiped && ara[i] - 1 > ara[i - 1]) {
        skiped = true;
        finish = ara[i] - 1;
        start = ara[i - 1] + 1;
      } else if (skiped && ara[i] - 1 > ara[i - 1]) {
        done = true;
      }
    }
    if (done) {
      cout << -1 << endl;
      break;
    }
    queue<int> q;
    for (i = start; i <= finish; i++) {
      q.push(i);
    }
    vector<int> ans;
    int last;
    if (ara[0] == 1) {
      ans.push_back(0);
      last = 1;
    } else {
      ans.push_back(1);
      last = 0;
    }
    for (i = 1; i < n; i++) {
      if (ara[i] == ara[i - 1]) {
        ans.push_back(q.front());
        if (q.size() != 1) q.pop();
      } else {
        ans.push_back(last);
      }
      last = ara[i];
    }
    {
      if (1) {
        for (auto a : ans) cout << a << " ";
        cout << endl;
      }
    };
  }
  return 0;
}
