// O(N^2)解法
#include <bits/stdc++.h>
using namespace std;

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

using ll = long long;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  
  vector<ll> odd, even;
  for (int i = 0; i < n; i++) {
    ll a;
    cin >> a;
    if (a % 2 == 1) odd.emplace_back(a);
    else even.emplace_back(a);
  }
  
  sort(odd.begin(), odd.end());
  sort(even.begin(), even.end());
  
  // sentinel
  odd.emplace_back(2e18 - 1);
  even.emplace_back(2e18);
  
  ll ans = n;
  int even_it = 0;
  for (int i = 0; i < odd.size() - 1; i++) {
    while (even.at(even_it) <= odd.at(i)) even_it++;
    if (even.at(even_it) == odd.at(i) + 1) {
      ll keep = 1, it2 = even_it;
      while (true) {
        if (even.at(it2) + 2 == even.at(it2 + 1)) {
          keep++;
          it2++;
        }
        else {
          break;
        }
      }
      ans += keep;
    }
  }
  int odd_it = 0;
  for (int i = 0; i < even.size() - 1; i++) {
    while (odd.at(odd_it) <= even.at(i)) odd_it++;
    if (odd.at(odd_it) == even.at(i) + 1) {
      ll keep = 1, it2 = odd_it;
      while (true) {
        if (odd.at(it2) + 2 == odd.at(it2 + 1)) {
          keep++;
          it2++;
        }
        else {
          break;
        }
      }
      ans += keep;
    }
  }
  
  cout << ans << '\n';
}