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 MX = 10010896;
const int lmt = 3164;
const int N = 10000001;
const int MOD = 1000000007;
using namespace std;
long long int extgcd(long long int a, long long int b, long long int& x,
                     long long int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    int g = extgcd(b, a % b, y, x);
    y -= a / b * x;
    return g;
  }
}
long long int modpow(long long int a, long long int b) {
  long long int res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
inline void readi(int& n) {
  n = 0;
  char c = getchar();
  while (c < '0' or c > '9') c = getchar();
  while (c >= '0' and c <= '9') {
    n = (n << 3) + (n << 1) + c - '0';
    c = getchar();
  }
}
inline void readl(long long int& n) {
  n = 0;
  char c = getchar();
  while (c < '0' or c > '9') c = getchar();
  while (c >= '0' and c <= '9') {
    n = (n << 3) + (n << 1) + c - '0';
    c = getchar();
  }
}
inline void print(long long int a) {
  char s[20];
  int i = 0;
  do {
    s[i++] = a % 10 + '0';
    a /= 10;
  } while (a);
  i--;
  while (i >= 0) putchar(s[i--]);
  putchar(' ');
}
inline void printl(long long int a) {
  char s[20];
  int i = 0;
  do {
    s[i++] = a % 10 + '0';
    a /= 10;
  } while (a);
  i--;
  while (i >= 0) putchar(s[i--]);
  putchar('\n');
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int* a = new int[n];
  for (int i = (int)0; i < (int)n; ++i) cin >> a[i];
  int flag = 1;
  for (int i = (int)0; i < (int)n; ++i) {
    if (a[i] > (i + 1)) {
      flag = 0;
    }
  }
  if (!flag)
    cout << -1;
  else {
    int* ans = new int[n];
    for (int i = (int)0; i < (int)n; ++i) ans[i] = -1;
    set<int> found;
    for (int i = (int)n - 1; i >= (int)0; --i) {
      if (found.find(a[i]) == found.end()) {
        if (i != n - 1) ans[i + 1] = a[i];
        found.insert(a[i]);
      }
    }
    int hold = 0;
    for (int i = (int)0; i < (int)n; ++i) {
      while (found.find(hold) != found.end()) hold++;
      if (ans[i] != -1) continue;
      ans[i] = hold;
      hold++;
    }
    for (int i = (int)0; i < (int)n; ++i) cout << ans[i] << " ";
  }
}
