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;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const int mod = (int)1e9 + 7;
const long long INF64 = 3e18;
void smxl(long long &a, long long b) {
  if (a < b) a = b;
}
void smnl(long long &a, long long b) {
  if (a > b) a = b;
}
void adsl(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void misl(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
void smx(long long &a, long long b) {
  if (a < b) a = b;
}
void smn(long long &a, long long b) {
  if (a > b) a = b;
}
void ads(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void mis(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long egcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long mbinp(long long a, long long b) {
  a %= mod;
  if (b == 0) return 1;
  long long ans = mbinp(a, b / 2);
  long long tmp = (ans * ans) % mod;
  if (b % 2) return ((tmp * a) % mod);
  return ((tmp) % mod);
}
long long binp(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = binp(a, b / 2);
  long long tmp = (ans * ans);
  if (b % 2) return ((tmp * a));
  return ((tmp));
}
long long C(int n, int m) {
  long long ret = 1;
  for (int i = 1; i <= m; i++) {
    ret *= (n - i + 1);
    ret /= i;
  }
  return ret;
}
long long overbinp(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      if (res < INF64 / a)
        res *= a;
      else
        return INF64;
    }
    if (b > 1) {
      if (a < INF64 / a)
        a *= a;
      else
        return INF64;
    }
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> all(n + 1, 0);
  for (int i = 1; i < n + 1; i++) cin >> all[i];
  vector<int> b(n + 1, -1);
  vector<bool> used(n, 0);
  for (int i = 1; i < n + 1; i++)
    if (all[i] != all[i - 1]) {
      b[i] = all[i - 1];
      used[all[i - 1]] = 1;
    }
  used[all[n]] = 1;
  int m = 0;
  for (int i = 1; i < n + 1; i++) {
    while (used[m]) m++;
    used[m] = 1;
    cout << (b[i] == -1 ? m : b[i]) << " ";
  }
}
