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 N = 1e5 + 5;
int last[N];
int a[N];
int b[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  bool ok = 1;
  int mx = -1;
  set<int> st;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] < mx || a[i] > i) ok = 0;
    mx = max(mx, a[i]);
    last[a[i]] = i;
    st.insert(i);
  }
  if (!ok) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    int t = last[i];
    if (t == 0) {
      auto it = st.begin();
      int id = *it;
      st.erase(id);
      b[id] = i;
    } else {
      b[t + 1] = i;
      st.erase(t + 1);
    }
  }
  if (st.size()) {
    auto it = st.begin();
    for (; it != st.end(); ++it) {
      b[*it] = n;
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << b[i + 1] << ' ';
  }
}
