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;
using cd = complex<double>;
const double PI = acos(-1);
void fft(vector<cd>& a, bool invert) {
  int n = a.size();
  if (n == 1) return;
  vector<cd> a0(n / 2), a1(n / 2);
  for (int i = 0; 2 * i < n; i++) {
    a0[i] = a[2 * i];
    a1[i] = a[2 * i + 1];
  }
  fft(a0, invert);
  fft(a1, invert);
  double ang = 2 * PI / n * (invert ? -1 : 1);
  cd w(1), wn(cos(ang), sin(ang));
  for (int i = 0; 2 * i < n; i++) {
    a[i] = a0[i] + w * a1[i];
    a[i + n / 2] = a0[i] - w * a1[i];
    if (invert) {
      a[i] /= 2;
      a[i + n / 2] /= 2;
    }
    w *= wn;
  }
}
vector<int> multiply(vector<int> const& a, vector<int> const& b) {
  vector<cd> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  int n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  fa.resize(n);
  fb.resize(n);
  fft(fa, false);
  fft(fb, false);
  for (int i = 0; i < n; i++) fa[i] *= fb[i];
  fft(fa, true);
  vector<int> result(n);
  for (int i = 0; i < n; i++) result[i] = round(fa[i].real());
  return result;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool pali(string s) {
  long long l = s.length();
  for (int i = 0; i < l / 2; i++) {
    if (s[i] != s[l - i - 1]) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  long long* a = new long long[n];
  long long maxi = -1;
  long long ans = true;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < maxi) {
      ans = false;
    }
    maxi = max(maxi, a[i]);
  }
  if (ans) {
    long long val = n;
    vector<long long> empty;
    vector<long long> mex;
    bool ans = true;
    mex.push_back(-1);
    long long* b = new long long[n];
    for (int i = 0; i < n; i++) {
      b[i] = -1;
    }
    if (a[0] == 0)
      empty.push_back(0);
    else {
      mex.push_back(0);
      b[0] = 0;
    }
    for (int i = 1; i < n; i++) {
      if (a[i] == a[i - 1]) {
        empty.push_back(i);
        continue;
      } else {
        long long p = a[i];
        empty.push_back(i);
        long long me = mex[mex.size() - 1] + 1;
        long long val = a[i] - 1;
        for (int j = me; j <= val; j++) {
          if (empty.size() == 0) {
            ans = false;
            break;
          }
          int l = empty[empty.size() - 1];
          b[l] = j;
          empty.pop_back();
        }
        if (ans) mex.push_back(val);
      }
      if (!ans) break;
    }
    if (!ans)
      cout << -1 << endl;
    else {
      long long tar = n;
      for (int i = 0; i < n; i++) {
        if (b[i] == -1) {
          b[i] = tar;
          tar--;
        }
      }
      for (int i = 0; i < n; i++) cout << b[i] << ' ';
    }
  } else
    cout << -1 << endl;
}
