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 long long MOD = 1e9 + 7;
const long long INF = 1e18;
const long double E = 1e-12;
const long double PI = acos(-1);
mt19937_64 rnd(time(0));
long long p1 = 31, p2 = 37, p3 = 7, p4 = rnd() % 100 + 1, md1 = MOD,
          md2 = 998244353, md3 = 998244357, md4 = rnd() % 1000000000;
long long n, m, k, h, T, kl = 0, sz = 0, a[1000001], b[1000001];
vector<long long> v[1000001];
long long binpow(long long h, long long r, long long md = MOD) {
  long long l = 1;
  while (r) {
    if (r & 1) l *= h, l %= md;
    h *= h;
    h %= md;
    r /= 2;
  }
  return l;
}
long long gcd(long long x, long long y) {
  if (x < y) swap(x, y);
  while (x && y) {
    x %= y;
    swap(x, y);
  }
  return x + y;
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) b[i] = -1;
  a[0] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] != a[i - 1]) {
      for (int j = a[i - 1] + 1; j <= a[i] - 1; j++) v[i].push_back(j);
      b[i] = a[i - 1];
    }
  }
  vector<long long> w;
  for (int i = n; i > 0; i--) {
    for (int j = 0; j < v[i].size(); j++) w.push_back(v[i][j]);
    if (b[i] != -1) continue;
    if (w.size() == 0)
      b[i] = n + 1;
    else
      b[i] = w.back(), w.pop_back();
  }
  if (w.size())
    cout << "-1\n";
  else {
    for (int i = 1; i <= n; i++) cout << b[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
