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 double PI = acos(-1);
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
int _runtimeTerror_() {
  long long n;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  long long ans[n];
  memset(ans, -1, sizeof(ans));
  int cur = 0, p = 0;
  bool mark[1000005];
  memset(mark, false, sizeof(mark));
  if (a[0] == 1) {
    ans[0] = 0, ++cur;
    mark[0] = true;
    mark[1] = true;
  } else {
    mark[0] = true;
  }
  for (int i = 1; i < n; ++i) {
    if (a[i] != a[i - 1]) {
      ans[i] = a[i - 1];
      mark[a[i - 1]] = true;
      mark[a[i]] = true;
    } else {
      ++p;
    }
  }
  long long curp = 0;
  for (int i = 0; i < n; ++i) {
    if (ans[i] == -1) {
      while (mark[curp]) ++curp;
      ans[i] = curp;
      ++curp;
    }
  }
  for (int i = 0; i < n; ++i) cout << ans[i] << " ";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  cerr << "\n" << 1.0 * clock() / CLOCKS_PER_SEC;
  return 0;
}
