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;
using ii = pair<int, int>;
using lint = long long;
const lint LINF = 1e18;
const int INF = 1e9 + 7;
const int mxn = 100010;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
int tc, N, M, K, a[mxn], b[mxn];
vector<vector<int>> adj;
set<int> st;
void input() {
  st.clear();
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> a[i];
  for (int i = 1; i < a[N]; i++) st.insert(i);
}
lint solve() {
  lint ret = 0;
  for (int i = N; i >= 2; i--) {
    if (a[i] != a[i - 1]) {
      b[i] = a[i - 1];
      st.erase(a[i - 1]);
    } else {
      if (st.empty()) {
        b[i] = 1000000;
        continue;
      }
      auto it = st.rbegin();
      if (*it > a[i]) {
        b[i] = *it;
        st.erase(*it);
      }
    }
  }
  if (a[1] != 0)
    b[1] = 0;
  else
    b[1] = 1;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  tc = 1;
  while (tc--) {
    input();
    lint ans = solve();
    for (int i = 1; i <= N; i++) cout << b[i] << ' ';
    cout << '\n';
  }
  return 0;
}
