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 + 5;
const int inf = 0x3f3f3f3f;
int b[maxn];
int a[maxn];
int ma[maxn];
int pre[maxn * 10];
int _find(int x) { return x == pre[x] ? x : pre[x] = _find(pre[x]); }
void _union(int x, int y) {
  x = _find(x), y = _find(y);
  if (x < y)
    pre[x] = y;
  else
    pre[y] = x;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < maxn * 10; ++i) pre[i] = i;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    ma[i] = max(ma[i - 1], a[i]);
  }
  bool flag = true;
  int x = 0;
  for (int i = n; i > 0; --i) {
    if (a[i] > ma[i - 1])
      b[i] = ma[i - 1];
    else
      b[i] = a[i] + 1;
    b[i] = _find(b[i]);
    _union(b[i], b[i] + 1);
    _union(a[i], a[i] + 1);
    x = max(x, a[i]);
  }
  if (!flag)
    puts("-1");
  else {
    for (int i = 1; i <= n; ++i) printf("%d%c", b[i], i == n ? '\n' : ' ');
  }
  return 0;
}
