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 unsigned long long mod = 998244353;
const long long inf = ((1ll << 31ll) - 1ll);
const long long INF = (((1ll << 59ll) - 1ll) * 2ll) + 1ll;
const long double pi = acos(-1);
long long qpow(long long b, long long e) {
  if (!e) return 1;
  if (e & 1) return qpow(b, e - 1) * b % mod;
  long long pwur = qpow(b, e >> 1);
  return pwur * pwur % mod;
}
long long t, n, x, m, lB, rB, B[200007], A[200007];
bool Flag[100006];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(0);
  srand(time(NULL));
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> B[i];
    Flag[B[i]] = true;
  }
  long long cont = 0;
  for (long long i = 1; i <= n; i++) {
    if (B[i - 1] != B[i]) {
      cout << B[i - 1];
    } else {
      do {
        cont++;
      } while (Flag[cont]);
      cout << cont;
    }
    cout << ' ';
  }
  return 0;
}
