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 N = 100005;
int a[N];
int vis[N];
int b[N];
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    vis[a[i]]++;
    if (a[i] != a[i - 1]) {
      b[i] = a[i - 1];
    }
    if (a[i] > i) {
      puts("-1");
      return;
    }
  }
  int t = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i] != 0) {
      continue;
    }
    while (vis[t] != 0 && t < 100005) {
      t++;
    }
    b[i] = t;
    t++;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] != 0 && a[i - 1] == 0) {
      b[i] = 0;
    }
    printf("%d ", b[i]);
  }
  cout << endl;
}
int main() {
  solve();
  return 0;
}
