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;
vector<string> vec_splitter(string s) {
  for (char& c : s) c = c == ',' ? ' ' : c;
  stringstream ss;
  ss << s;
  vector<string> res;
  for (string z; ss >> z; res.push_back(z))
    ;
  return res;
}
void debug_out(vector<string> args, int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
  if (idx > 0) cerr << ", ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, T...);
}
void localTest() {}
const long long N = 1e5 + 5;
long long A[N], B[N], par[N], mx[N], mn[N];
long long find(long long x) {
  if (x == par[x]) return x;
  return par[x] = find(par[x]);
}
void merge(long long x, long long y) {
  if (x < y) return;
  long long rx = find(x), ry = find(y);
  if (rx != ry) {
    par[rx] = ry;
    mn[ry] = min(mn[ry], ry);
    mx[ry] = max(mx[ry], ry);
  }
  return;
}
int main() {
  localTest();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, hi;
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> A[i];
  hi = A[n] - 1;
  set<long long> st;
  vector<long long> ans;
  for (long long i = 0; i <= hi; ++i) st.insert(i);
  for (long long i = 1; i <= n; ++i) {
    if (A[i]) {
      auto pp = st.lower_bound(A[i] - 1);
      if (pp == st.end()) {
        if (!st.empty()) {
          ans.push_back(*st.begin());
          st.erase(st.begin());
        } else {
          ans.push_back(A[i] - 1);
        }
        continue;
      }
      if (*pp == A[i]) {
        if (pp == st.begin()) {
          ++pp;
          if (pp == st.end()) {
            cout << -1 << "\n";
            return 0;
          }
          ans.push_back(*pp);
          st.erase(pp);
          continue;
        }
        --pp;
        ans.push_back(*pp);
        st.erase(pp);
      } else {
        ans.push_back(*pp);
        st.erase(pp);
      }
    } else {
      auto pp = st.lower_bound(1LL);
      if (pp == st.end()) {
        ans.push_back(1);
        continue;
      } else {
        ans.push_back(*pp);
        st.erase(pp);
      }
    }
  }
  for (auto u : ans) cout << u << " ";
  cout << "\n";
  return 0;
}
