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;
long long int power(long long int x, long long int y, long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long int nCr(long long int n, long long int r, long long int m) {
  if (r > n) return 0;
  long long int a = 1, b = 1, i;
  for (i = 0; i < r; i++) {
    a = (a * n) % m;
    --n;
  }
  while (r) {
    b = (b * r) % m;
    --r;
  }
  return (a * power(b, m - 2, m)) % m;
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  int f = 0;
  map<int, int> mp, mp2;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    mp[v[i]] = 1;
    if (i != 0 && v[i - 1] > v[i]) f = 1;
  }
  if (f || v[0] > 1) {
    cout << -1 << '\n';
    return;
  }
  vector<int> ans(n);
  int l = 0;
  while (mp[l] == 1) ++l;
  if (v[0] == 1) {
    ans[0] = l;
    l++;
  } else {
    ans[0] = l;
    ++l;
  }
  while (mp[l] == 1) ++l;
  for (int i = 1; i < n; ++i) {
    while (mp[l] == 1) ++l;
    if (v[i] != v[i - 1])
      ans[i] = v[i - 1];
    else
      ans[i] = l++;
    while (mp[l] == 1) ++l;
  }
  int mex = 0;
  for (int i = 0; i < n; ++i) {
    mp2[ans[i]] = 1;
    while (mp2[mex] == 1) mex++;
    if (mex != v[i]) f = 1;
  }
  if (f) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 0; i < n; ++i) cout << ans[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
