// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



int n;

int a[101010];

int b[101010];



int main()

{

  cin >> n;

  for (int i = 0; i < n; i++){

    cin >> a[i];

    b[a[i]]++;

  }



  int ans = 0;

  int po = 0;

  for (int i = 0; i < 101010; i++){

    if (b[i] & 1){

      ans++;

    } else if (b[i] == 0) {

     continue;

    }

    else {

      po++;

    }

  }



  cout << ans + po / 2 * 2 << endl;

}




// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;



// clang-format off

#define  range(i, l, r) for (int i = (int)(l); i < (int)(r); ++(i))

#define rrange(i, l, r) for (int i = (int)(r) - 1; i >= (int)(l); --(i))



#define  whole(f, x, ...) ([&](decltype((x)) container) { return (f)(  begin(container),  end(container), ## __VA_ARGS__); })(x)

#define rwhole(f, x, ...) ([&](decltype((x)) container) { return (f)( rbegin(container), rend(container), ## __VA_ARGS__); })(x)



#define endl '\n'

#define debug(x) cerr << "(" << __LINE__ << ")" << #x << ": " << (x) << endl



using i32 = int;

using u32 = unsigned int;

using i64 = long long;

using u64 = unsigned long long;



// constexpr i32 mod   = 998244353;

constexpr i32 mod   = 1e9 + 7;

constexpr i32 inf   = 1001001001;

constexpr i64 infll = 1001001001001001001ll;



constexpr int dx[] = {0, -1, 1, 0, -1, 1, -1, 1}; 

constexpr int dy[] = {-1, 0, 0, 1, -1, -1, 1, 1};



struct IoSetup { IoSetup(int x = 15){ cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(x); cerr << fixed << setprecision(x); } } iosetup;



template <typename T = i64> T input() { T x; cin >> x; return x; }



template <typename T> ostream &operator<<(ostream &os, vector<T> &v) { range(i, 0, v.size()) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } 

template <typename T> istream &operator>>(istream &is, vector<T> &v) { for (T &in : v) is >> in; return is; }



template <typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }

template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }



template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }

template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }



template <typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }

template <typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

// clang-format on



void solver() {

    int N = input();

    map< int, int > cnt;

    range(i, 0, N) {

        cnt[input()]++;

    }

    

//    vector< int > A(N);

//    cin >> A;

    

    int ans = 0;

    int even_cnt = 0;

//    for (auto a: A) even_cnt += !(a & 1);

    for (auto p: cnt) {

        ans++;

        even_cnt += !(p.second & 1);

    }

    

    ans -= (even_cnt & 1);



    cout << ans << endl;

}



signed main(int argc, char *argv[]) {

    solver();

}


// Fastest known solution for this problem: