// Original C++ solution (slower):
    /**



    +++---------+++--------- Don't lose faith ---------+++---------+++



    +++---------+++--------- Author : sea_26 ----------+++---------+++



    **/



    #pragma GCC optimize("03")

    #include <bits/stdc++.h>



    using namespace std;



    using ll = long long;

    using ld = long double;



    priority_queue<int> pq;



    const int M = 1e9 + 7;

    const ll MOD = 998244353;

    const ld eps = 1e-9;

    const ld PI = acos(-1.0);



    // const double pi = atan2(0, -1);

    // const double PI = 4 * atan(1);

    // const double pi = 2*cos(0.0);



    const int iinf = 0x3f3f3f3f;

    const ll linf = 0x3f3f3f3f3f3f3f3f;



    #define mp make_pair

    #define eb emplace_back

    #define ep emplace

    #define f first

    #define s second

    #define sz size()



    #define bp(n)  __builtin_popcount(n)

    #define bclz(n) __builtin_clzll(n)

    #define bctz(n) __builtin_ctzll(n)

    #define bip(n) __builtin_parity(n)



    #define vsort(v) sort(v.begin(),v.end())

    #define rvsort(v) sort(v.begin(),v.end(),greater<int>())

    #define mnv(v) *min_element(v.begin(),v.end())

    #define mxv(v) *max_element(v.begin(),v.end())

    #define all(v) (v).begin(),(v).end()

    #define ita(v) iota(all(v), 0)



    #define ub upper_bound

    #define lb lower_bound

    #define cr(n) (floor)cbrt(n)

    #define msz(a) memset(a, 0, sizeof(a))

    #define msm(a) memset(a, 0xff, sizeof(a))



    #define rep(n) for(int i=0; i<n; i++)

    #define rrep(n) for(int i=n-1; i>=0; i--)

    #define lrep(p,q) for(int j=p; j<=q; j++)

    #define ldrep(p,q) for(int i=p; i>=q; i--)

    #define rsrep(n) for(int i=2; i*i<=n; i++)

    #define it(v)  for(auto p: v)

    #define yes cout << "YES" << endl

    #define no cout << "NO" << endl



    #define fs ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);



    template< typename T> inline T MAX(T a, T b){ return a ^ ((a ^ b) & -(a < b)); }

    template< typename T> inline T MIN(T a, T b){ return b ^ ((a ^ b) & -(a < b)); }

    template< typename T> void tpns(T n){ rrep(30){ if(n & (1 << i)) pq.push(1 << i); } }

    template< typename T> void SWAP(T &a, T &b){ a = ((a) ^= (b)),  b = ((b) ^= (a)),  a = ((a) ^= (b)); }

    template< typename T> inline T mod(T a, T b){ return a >= b ? a % b : a; }

    template< typename T> inline T gcd(T a, T b){ while (b) b ^= a ^= b ^= a %= b; return a; }

    ///template< typename T> inline T gcd(T a, T b) {while(b){int r = a % b; a = b; b = r;}return a; } /// T.C : log(a+b)

    template< typename T> inline T lcm(T a, T b){ return (a / gcd(a,b)) * b; }

    template< typename T> T sqr(T x){ return x * x; }

    template< typename T> T bigmod(T n,ll p, T M){ if(!p) return 1; else{ T r = sqr(bigmod(n,p >> 1, M))%M; if(p&1) r = (r*n)%M; return r; }}

    template< typename T> T exgcd(T a, T b, T& x, T& y){ if(!b){ x = 1; y = 0;return a;} else{T g = exgcd(b,a%b,y,x); y -= (a/b)*x; return g;}}

    template< typename T> T icbrt(T mul){T r = cbrt(mul) + 1; while(r * r * r > mul) r--; return r; } /// 54229668



    ///ll gcdextd(ll a, ll b, ll& x, ll& y) { if (!a) { x = 0; y = 1; return b; }  ll y1,d; d = gcdextd(b % a, a, y, y1);  x = y1 - (b / a) * y; return d; }

    ll takemod(ll a){return ((a % M) + M) % M; }

    ll fexpo(ll base, ll e){ ll res = 1; while(e > 0){ if(e & 1) res = (res * base) % M; base = (base * base ) % M; e >>= 1; } return 0; }

    ll modinv(ll a){return takemod(fexpo(takemod(a), M-2));}



    void ReT(){ cout << "\nTime = " << fixed << setprecision(3) << double(clock()) / CLOCKS_PER_SEC  << " Sec"<< endl; exit(0);}







    int main(){



        fs;



        int n, x, uq;

        map<int,int> card;



        cin >> n;



        rep(n){

            cin >> x;

            card[x]++;

        }



        uq=card.sz;

        

        if(uq & 1) cout << uq << endl;

        else cout << uq-1 << endl;



        /// ReT();



    return 0;

    }


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

const int M = 1e5+5;



int arr[M] = {0};



int main(){

	int n,i,j;

	int num1 = 0;

	int num2 = 0;

	int ans = 0;

	scanf("%d",&n);

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

		scanf("%d",&j);

		if(arr[j] == 0){

			arr[j] = 1;

			num1++;

		}

		else{

			num2++;

		}

	}

	if(num2 % 2 == 0) ans = num1;

	else ans = num1 -1;

	printf("%d",ans);

}

// Fastest known solution for this problem: