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 double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
int n;
int a[maxn], b[maxn];
vector<int> ve;
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  if (a[1] != 0 && a[1] != 1) {
    puts("-1");
    return 0;
  }
  int las = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == las)
      ve.push_back(i);
    else {
      b[i] = las;
      int diff = a[i] - las - 1;
      if (ve.size() < diff) {
        puts("-1");
        return 0;
      }
      int tp = las + 1;
      while (diff--) {
        b[ve.back()] = tp++;
        ve.pop_back();
      }
      las = a[i];
    }
  }
  while (!ve.empty()) {
    b[ve.back()] = 1000000;
    ve.pop_back();
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", b[i]);
  }
  puts("");
  return 0;
}
