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 N = 150003;
const long long INF = 1e15 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long a[n + 3];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long b[n + 3];
  long long i = 0;
  long long mn;
  long long ans = 1;
  vector<long long> vi;
  if (a[0] == 1) {
    while (a[i] == 1) b[i] = 0, i++;
    mn = 0;
    vi.push_back(0);
  } else if (a[0] == 0) {
    while (a[i] == 0) b[i] = 1, i++;
    mn = 1;
    vi.push_back(0);
  } else {
    cout << "-1" << endl;
    return 0;
  }
  vector<long long> vm;
  for (long long j = i; j < n; j++) {
    if (a[j] == a[j - 1])
      b[j] = b[j - 1];
    else if (a[j] > a[j - 1]) {
      vm.push_back(a[j]);
      vi.push_back(j);
      b[j] = a[j - 1];
    } else {
      cout << "-1" << endl;
      ans = 0;
      break;
    }
  }
  if (ans == 0) return 0;
  map<long long, long long> m;
  m[0] = m[1] = 1;
  long long k = 0;
  for (long long i = 0; i < vm.size(); i++) {
    long long j = vi[k] + 1;
    for (long long p = a[vi[k]] + 1; p < vm[i]; p++) {
      if (m[p] == 1)
        continue;
      else {
        if (j == vi[k + 1]) {
          ans = 0;
          break;
        }
        b[j] = p;
        m[p] = 1;
        j++;
      }
    }
    k++;
    if (ans == 0) break;
  }
  if (ans == 0) {
    cout << "-1" << endl;
  } else {
    for (long long i = 0; i < n; i++) cout << b[i] << " ";
    cout << endl;
  }
}
