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;
bool vis[1000008];
int link[1000008];
void solve() {
  int a, k, i, j, n, sum = 0, mx = 0, ind, ct = 1, curr = 0, remain = -1;
  cin >> n;
  int D[n + 5];
  for (i = 0; i < n; i++) {
    cin >> D[i];
    vis[D[i]] = 1;
  }
  link[D[n - 1]] = D[n - 1];
  for (i = 0; i < n - 1; i++) {
    link[D[i]] = D[i + 1];
  }
  if (D[0] == 0) {
    while (vis[++curr])
      ;
    cout << curr << " ";
    while (vis[++curr])
      ;
    remain = 0;
  } else {
    cout << 0 << " ";
    while (vis[++curr])
      ;
  }
  for (i = 1; i < n; i++) {
    if (D[i] != D[i - 1]) {
      cout << D[i - 1] << " ";
      if (remain == D[i - 1]) remain = -1;
      continue;
    }
    if (remain != D[i] && remain > -1) {
      cout << remain << " ";
      remain = -1;
    } else {
      cout << curr << " ";
      while (vis[++curr])
        ;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
