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 MOD = 998244353;
const long long mod = 1e9 + 7;
const int mxN = 1e5 + 9;
int a[mxN];
int ans[mxN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] > i) {
      cout << -1;
      return 0;
    }
  }
  vector<int> free;
  int M = -1;
  for (int i = 1; i <= n; i++) {
    free.push_back(i);
    if (a[i] < i && (M + 1 <= a[i] - 1)) {
      for (int j = M + 1; j <= a[i] - 1; j++) {
        int tmp = free.back();
        free.pop_back();
        ans[tmp] = j;
      }
      M = a[i] - 1;
    } else if (a[i] == i) {
      for (int j = M + 1; j <= a[i] - 1; j++) {
        int tmp = free.back();
        free.pop_back();
        ans[tmp] = j;
      }
      M = a[i] - 1;
    }
  }
  int xx = a[n];
  for (auto &c : free) {
    ++xx;
    ans[c] = xx;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
