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 long long N = 1e6 + 10;
long long PI = acos(-1.0), MOD = 1000000007;
long long n, m, x, y, z, k, sol, sum, ans, l, r, xx, yy, t;
long long a[N], ma[N], mem[1000][1000];
vector<long long> v;
vector<long long> v1;
vector<long long> v2;
vector<pair<long long, long long> > vp;
map<long long, long long> b;
string s1, s2, s;
set<long long> st;
char c;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    x = max(a[i], x);
    ma[a[i]]++;
  }
  vector<long long> ans;
  for (int i = 0; i < 1e5 + 10; i++) {
    if (!ma[i]) st.insert(i);
  }
  for (long long i = 0; i < n; i++) {
    x = a[i];
    if (x > ans.size() + 1) {
      cout << -1 << endl;
      return 0;
    } else {
      auto it = st.begin();
      ans.push_back(*it);
      st.erase(it);
      ma[a[i]]--;
      if (ma[a[i]] == 0) {
        st.insert(a[i]);
      }
    }
  }
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
