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;
vector<long long> v;
long long arr[100005];
long long ans[100005];
map<long long, long long> m;
int main() {
  long long n;
  while (cin >> n) {
    for (long long i = 1; i <= n; i++) cin >> arr[i];
    memset(ans, -1, sizeof(ans));
    m.clear();
    if (arr[1] == 0) {
      ans[1] = 1;
      m[1] = 1;
    } else {
      ans[1] = 0;
      m[0] = 0;
    }
    if (arr[1] != 0 && arr[1] != 1) {
      cout << -1 << endl;
      continue;
    }
    for (long long i = 2; i <= n; i++) {
      if (arr[i] != arr[i - 1]) {
        m[arr[i - 1]] = arr[i - 1];
        ans[i] = arr[i - 1];
      }
    }
    long long x = 0;
    long long ck = 0;
    m[arr[n]] = arr[n];
    for (long long i = 2; i <= n; i++) {
      while (m.find(x) != m.end()) x++;
      if (ans[i] != -1) {
        if (x < (arr[i] - 1)) ck = 1;
      } else {
        ans[i] = x;
        x++;
      }
    }
    if (ck == 1)
      cout << -1 << endl;
    else {
      for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
      cout << endl;
    }
  }
}
