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;
long long a[1000005];
long long b[1000005];
long long tp[1000005];
long long pp[1000005];
long long cc[1000005];
vector<long long> V;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  int bj = 0;
  long long minn = 0;
  long long mi = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    tp[a[i]]++;
  }
  if (a[1] == 1) {
    b[1] = 0;
    pp[0]++;
    mi = 1;
  } else if (a[1] == 0) {
    while (tp[minn] > 0) {
      minn++;
    }
    b[1] = minn;
    pp[minn]++;
    mi = 0;
  }
  tp[a[1]]--;
  if (tp[a[1]] == 0) {
    minn = min(minn, a[1]);
  }
  for (int i = 2; i <= n; i++) {
    long long cnt = mi;
    if (mi >= a[i]) {
      cnt = b[i - 1] + 1;
      while (tp[cnt] > 0) {
        cnt++;
      }
      b[i] = cnt;
    } else {
      while (tp[cnt] > 0) {
        cnt++;
      }
      b[i] = cnt;
      pp[cnt]++;
      while (pp[mi] > 0) {
        mi++;
      }
    }
    tp[a[i]]--;
    if (tp[a[i]] == 0) {
      mi = min(mi, a[i]);
    }
  }
  long long M = 0;
  for (int i = 1; i <= n; i++) {
    cc[b[i]]++;
    while (cc[M] > 0) {
      M++;
    }
    if (M < a[i]) {
      bj = 1;
      break;
    }
  }
  if (bj == 1) {
    cout << "-1\n";
  } else {
    for (int i = 1; i <= n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}
