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 NN = 1e7 + 7;
const int N = 5e2 + 7;
const int M = 26;
const long long mod = 998244353;
const long long inf = 1e18 + 7;
const double long rf = 1e-14;
const int B = sqrt(N);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void solve1() {
  int n;
  cin >> n;
  vector<int> a(n, 0);
  vector<pair<int, int> > b;
  for (auto &x : a) cin >> x;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j < n && a[i] == a[j]) j += 1;
    j -= 1;
    b.push_back({j, a[j]});
    i = j;
  }
  set<int> s;
  for (int i = 1; i < n; i++) {
    for (int j = a[i - 1] + 1; j < a[i]; j++) {
      s.insert(j);
    }
  }
  vector<int> ans;
  int mn = 0;
  vector<int> used(n + 1, 0);
  for (int i = 0; i < a[0]; i++) s.insert(i);
  for (int i = a[n - 1] + 1; i < a[n - 1] + n + 1; i++) s.insert(i);
  for (int i = 0, j = 0; i < n; i++) {
    while (j < (int)b.size() && b[j].first < i) {
      s.insert(b[j].second);
      j += 1;
    }
    ans.push_back(*s.begin());
    s.erase(s.begin());
    used[ans.back()] = 1;
    while (used[mn]) mn += 1;
    if (mn != a[i]) {
      cout << i << " ";
      cout << "-1\n";
      return;
    }
  }
  for (auto x : ans) cout << x << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int cghf = 1;
  while (cghf--) {
    solve1();
  }
}
