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 main() {
  int n, a[100005];
  while (scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    int b[n + 2];
    memset(b, -1, sizeof(b));
    set<int> SET;
    for (int i = 0; i < a[n - 1]; i++) SET.insert(i);
    for (int i = 0; i < n; i++) {
      if (i == 0) {
        if (a[i] > 0) b[i] = 0, SET.erase(0);
      } else if (a[i - 1] < a[i])
        b[i] = a[i - 1], SET.erase(a[i - 1]);
    }
    for (int i = 0; i < n; i++) {
      if (b[i] == -1) {
        if (SET.size() == 0)
          b[i] = n + 1;
        else {
          b[i] = *SET.begin();
          SET.erase(SET.begin());
        }
      }
    }
    for (int i = 0; i < n; i++) {
      printf("%d ", b[i]);
    }
    printf("\n");
    SET.clear();
  }
  cerr << "\nTime : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec.\n";
  return 0;
}
