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> tree[2 * 400005], a(400005);
void build(long long n, long long b, long long e) {
  if (b == e) {
    tree[n].push_back(a[b]);
    return;
  }
  long long mid = (b + e) / 2;
  build(2 * n, b, mid);
  build(2 * n + 1, mid + 1, e);
  merge(tree[2 * n].begin(), tree[2 * n].end(), tree[2 * n + 1].begin(),
        tree[2 * n + 1].end(), back_inserter(tree[n]));
}
bool query(long long n, long long b, long long e, long long l, long long r,
           long long val) {
  if (r < b || e < l) return 0;
  if (l <= b && e <= r) {
    return binary_search(tree[n].begin(), tree[n].end(), val);
  }
  long long mid = (b + e) / 2;
  return query(2 * n, b, mid, l, r, val) |
         query(2 * n + 1, mid + 1, e, l, r, val);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, 1, n);
  vector<long long> ans;
  map<int, int> mp;
  long long now = 0;
  long long cur = 0;
  for (int i = 1; i <= n; i++) {
    long long need = a[i];
    while (query(1, 1, n, i, n, now)) {
      now++;
    }
    ans.push_back(now);
    now = need;
  }
  for (long long x : ans) {
    cout << x << " ";
  }
  return 0;
}
