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;
template <typename T>
string tos(T a) {
  stringstream ss;
  string ret;
  ss << a;
  ss >> ret;
  return ret;
}
void read(int &_a) { int temp = scanf("%d", &_a); }
void read(int &_a, int &_b) { int temp = scanf("%d %d", &_a, &_b); }
void read(int &_a, int &_b, int &_c) {
  int temp = scanf("%d %d %d", &_a, &_b, &_c);
}
const int SZ = 1e5 + 7;
int N, AR[SZ], ANS[SZ];
set<int> ST, DN;
bool validate() {
  ST.clear();
  for (int(x) = (0); (x) < (2 * N); (x)++) ST.insert(x);
  for (int(x) = (0); (x) < (N); (x)++) {
    if (ST.count(ANS[x])) ST.erase(ST.find(ANS[x]));
    if (*(ST.begin()) != AR[x]) return false;
  }
  return true;
}
int main() {
  while (cin >> N) {
    bool ok = true;
    ST.clear();
    DN.clear();
    for (int(x) = (0); (x) < (N); (x)++) {
      read(AR[x]);
      if (AR[x] > x + 1) ok = false;
    }
    if (!ok) {
      puts("-1");
      continue;
    }
    for (int(x) = (0); (x) < (N); (x)++) {
      if (!x) {
        if (AR[x])
          ANS[x] = 0;
        else
          ANS[x] = 1;
        DN.insert(ANS[x]);
      } else {
        if (AR[x] == AR[x - 1])
          ST.insert(x);
        else {
          ANS[x] = AR[x - 1];
          DN.insert(ANS[x]);
          for (int(i) = (AR[x - 1] + 1); (i) < (AR[x]); (i)++) {
            if (DN.count(i)) continue;
            if (ST.size() == 0) {
              ok = false;
              break;
            }
            int pos = *(ST.begin());
            ST.erase(ST.begin());
            ANS[pos] = i;
            DN.insert(i);
          }
        }
      }
    }
    int vl = N + 7;
    for (__typeof__(ST.begin()) it = ST.begin(); it != ST.end(); it++) {
      ANS[*it] = vl++;
    }
    if (!ok || validate() == false) {
      puts("-1");
      continue;
    }
    for (int(x) = (0); (x) < (N); (x)++) {
      if (x) printf(" ");
      printf("%d", ANS[x]);
    }
    puts("");
  }
  return 0;
}
