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 int mod = 1000000007;
long long qpow(long long n, long long k) {
  long long ans = 1;
  assert(k >= 0);
  n %= mod;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k >>= 1;
  }
  return ans % mod;
}
int main() {
  int n;
  cin >> n;
  int arr[n + 1];
  arr[0] = -1;
  set<int> avl;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    for (int j = arr[i - 1] + 1; j < arr[i]; j++) {
      avl.insert(j);
    }
  }
  set<int> inl;
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (arr[i] == 0) {
      if (!avl.empty()) {
        ans.push_back(*avl.begin());
        inl.insert(*avl.begin());
        avl.erase(avl.begin());
      } else
        ans.push_back(arr[n] + 1);
    } else {
      int cur = arr[i];
      cur--;
      if (inl.find(cur) != inl.end()) {
        if (!avl.empty()) {
          ans.push_back(*avl.begin());
          inl.insert(*avl.begin());
          avl.erase(avl.begin());
        } else
          ans.push_back(arr[n] + 1);
      } else {
        auto it = avl.find(cur);
        if (it != avl.end()) avl.erase(it);
        ans.push_back(cur);
        inl.insert(cur);
      }
    }
    if (i < n && arr[i + 1] != arr[i]) avl.insert(arr[i]);
  }
  auto it = avl.find(arr[n]);
  if (it != avl.end()) avl.erase(it);
  if (avl.empty()) {
    for (auto it = ans.begin(); it != ans.end(); it++) {
      cout << *it << ' ';
    }
  } else
    cout << "-1";
  cout << '\n';
  return 0;
}
