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>
#pragma GCC optimize( \
    "Ofast,no-stack-protector,unroll-loops,no-stack-protector,fast-math")
using namespace std;
const int N = 1e5 + 5, M = 2 * N + 5;
bool vis[N];
int n, a[N], ans[N], idx[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    idx[a[i]] = i;
  }
  memset(ans, -1, sizeof ans);
  set<int> st;
  for (int i = 1; i <= n; ++i) st.emplace(i);
  int mx = *max_element(a + 1, a + n + 1);
  for (int i = 0; i < mx; ++i) {
    int here = idx[i] + 1;
    auto it = st.lower_bound(here);
    if (it == st.end()) exit(puts("-1"));
    here = *it;
    ans[here] = i;
    st.erase(here);
  }
  for (int i = 1; i <= n; ++i)
    if (ans[i] == -1) ans[i] = 1e5 + 1;
  int tmp = 0;
  for (int i = 1; i <= n; ++i) {
    vis[ans[i]] = 1;
    while (tmp <= mx && vis[tmp]) ++tmp;
    if (tmp > mx) exit(puts("-1"));
    if (a[i] != tmp) exit(puts("-1"));
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
}
