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 maxn = 1e5 + 10;
int n, A[maxn];
map<int, int> vis;
vector<int> ans;
void init() {
  vis.clear();
  ans.clear();
}
void solve() {
  int fst = 0;
  for (int i = (1); i <= (n); i++) {
    if (A[i] > i + 1) {
      printf("-1\n");
      return;
    }
    if (i > 1 && A[i] > A[i - 1]) {
      ans.push_back(A[i - 1]);
      continue;
    }
    while (vis.find(fst) != vis.end()) fst++;
    ans.push_back(fst);
    fst++;
  }
  for (auto x : ans) printf("%d ", x);
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &A[i]);
    vis[A[i]]++;
  }
  solve();
}
