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 MAXN = 5000001;
const long long int mod = 998244353;
const long long int INF = INT_MAX;
inline long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
inline long long int pow(long long int a, long long int n, long long int mod) {
  if (n == 0) return 1;
  long long int p = pow(a, n / 2, mod);
  p = (p % mod * p % mod) % mod;
  if (n % 2)
    return (p % mod * a % mod) % mod;
  else
    return p;
}
vector<long long int> prime(10000005, 1);
void sieve(long long int n) {
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == 1) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = 0;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    set<long long int> s;
    long long int mx = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      mx = max(mx, a[i]);
    }
    vector<long long int> ans(n);
    ans[0] = 0;
    for (long long int i = 1; i < n; i++) ans[i] = a[i - 1];
    for (long long int i = 1; i <= mx; i++) s.insert(i);
    for (long long int i = 0; i < n; i++) {
      s.erase(a[i]);
    }
    for (long long int i = n - 1; i >= 0; i--) {
      if (s.empty()) break;
      if (ans[i] == a[i] || a[i] == 0) {
        auto it = s.begin();
        ans[i] = *(s.begin());
        s.erase(it);
      }
    }
    for (long long int i = 0; i < n; i++) {
      if (ans[i] == 0 && a[i] == 0) ans[i] = mx + 1;
    }
    for (long long int i = 0; i < n; i++) cout << ans[i] << " ";
  }
}
