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;
long long a[1000005];
long long b[1000005];
long long tp[5000005];
long long pp[1000005];
long long cc[1000005];
vector<long long> V;
priority_queue<long long, vector<long long>, greater<long long>> Q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  int bj = 0;
  long long minn = 0;
  long long mi = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    tp[a[i]]++;
  }
  for (long long i = 0; i <= 3 * n; i++) {
    if (tp[i] == 0) {
      Q.push(i);
    }
  }
  long long cnt = 0;
  for (int i = 1; i <= n; i++) {
    b[i] = Q.top();
    Q.pop();
    tp[a[i]]--;
    if (tp[a[i]] == 0) {
      Q.push(a[i]);
    }
  }
  long long M = 0;
  for (int i = 1; i <= n; i++) {
    cc[b[i]]++;
    while (cc[M] > 0) {
      M++;
    }
    if (M < a[i]) {
      bj = 1;
      break;
    }
  }
  if (bj == 1) {
    cout << "-1\n";
  } else {
    for (int i = 1; i <= n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}
