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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const int mod = 1e9 + 7;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int pct(int x) { return __builtin_popcount(x); }
int bit(int x) { return 31 - __builtin_clz(x); }
int cdiv(int a, int b) { return a / b + !(a < 0 || a % b == 0); }
void DBG() { cerr << "]" << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
const int maxn = 1e5 * 3 + 20;
int n, _, q, k;
int a[maxn];
int b[maxn];
void solve() {
  cin >> n;
  set<int> seen;
  for (signed i = 1; i <= n; i++) cin >> a[i], seen.insert(a[i]);
  set<int> has;
  set<int> spe;
  seen.insert(0);
  for (signed i = 0; i <= a[n]; i++) {
    if (i != a[n]) {
      has.insert(i);
    }
    if (!seen.count(i)) {
      spe.insert(i);
    }
  }
  int cnt = 1e5 + 2;
  while ((signed)has.size() < n) {
    has.insert(cnt);
    spe.insert(cnt);
    cnt++;
  }
  bool valid = true;
  for (signed i = 1; i <= n; i++) {
    if (a[i] == a[i - 1]) {
      int large = *(spe.begin());
      b[i] = large;
      has.erase(large);
      spe.erase(large);
    } else {
      b[i] = a[i - 1];
      if (!has.count(a[i - 1])) {
        valid = false;
        break;
      }
      has.erase(a[i - 1]);
    }
  }
  if (!valid) {
    cout << -1 << endl;
    exit(0);
  }
  for (signed i = 1; i <= n; i++) {
    cout << b[i] << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}
