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;
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
const long long maxn = 1e5 + 10, mod = 1e9 + 7, INF = 1e15, lg = 21;
long long a[maxn], b[maxn];
deque<int> q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  bool is = 0;
  cin >> n;
  long long maxi = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i != n - 1) {
      maxi = max(maxi, a[i]);
    }
  }
  if (a[n - 1] == 2 && maxi == 0) {
    for (int i = 0; i < n - 1; ++i) {
      cout << 1 << ' ';
    }
    cout << 0 << ' ';
    return 0;
  }
  if (a[0] == 1) {
    b[0] = 0;
  } else {
    b[0] = 1;
  }
  for (int i = 0; i < n - 1; ++i) {
    if (a[i + 1] == a[i] && a[i + 1] != 0) {
      b[i + 1] = 0;
      q.push_front(i + 1);
    } else if (a[i + 1] == a[i] && a[i + 1] == 0) {
      b[i + 1] = 1;
    } else if (a[i + 1] - a[i] == 1) {
      b[i + 1] = a[i + 1] - 1;
    } else {
      q.push_front(i + 1);
      long long delta = a[i + 1] - a[i];
      long long val = 1;
      while (delta && !q.empty()) {
        long long v = q.back();
        b[v] = a[i + 1] - val;
        val++;
        delta--;
        q.pop_back();
      }
      if (delta - 1 > 0) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << b[i] << ' ';
  }
}
