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 df = (int)(1e6);
int v[100000 + 5], sol[100000 + 5], fv[10 * 100000 + 5];
set<int> todo;
priority_queue<int> dorinte;
void nu() {
  cout << -1;
  exit(0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int i, j, z;
  for (i = 1; i <= n; i++) cin >> v[i];
  for (z = 0; z < n; z++) todo.insert(z);
  for (i = n; i > 1; i--) {
    if (v[i] == v[i - 1]) {
      if (!dorinte.empty()) {
        sol[i] = dorinte.top();
        dorinte.pop();
      } else
        sol[i] = df;
    } else {
      sol[i] = v[i - 1];
      auto it3 = todo.find(sol[i]);
      if (it3 != todo.end()) todo.erase(it3);
    }
    auto it = todo.begin(), it2 = todo.lower_bound(v[i]);
    while (it != it2) {
      dorinte.push(*it);
      it = todo.erase(it);
    }
  }
  if (v[1] > 1) nu();
  if (v[1] == 1)
    sol[1] = 0;
  else {
    if (!dorinte.empty())
      sol[1] = dorinte.top();
    else
      sol[1] = df;
  }
  int mex = 0;
  for (i = 1; i <= n; i++) {
    fv[sol[i]]++;
    if (mex == sol[i]) do {
        mex++;
      } while (fv[mex] != 0);
    if (mex != v[i]) nu();
  }
  for (i = 1; i <= n; i++) cout << sol[i] << ' ';
  return 0;
}
