Problem:
Score : 400 points

Problem Statement

Snuke has decided to play a game using cards.
He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.

He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.

Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.

Constraints

3 ≦ N ≦ 10^{5}

N is odd.

1 ≦ A_i ≦ 10^{5}

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_{N}

Output

Print the answer.

Sample Input 1

5
1 2 1 3 7

Sample Output 1

3

One optimal solution is to perform the operation once, taking out two cards with 1 and one card with 2. One card with 1 and another with 2 will be eaten, and the remaining card with 1 will be returned to deck. Then, the values written on the remaining cards in the deck will be pairwise distinct: 1, 3 and 7.

Sample Input 2

15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1

Sample Output 2

7

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
void sort(int* a, int N){
	int i,j;
	int mini;
	int temp;
	for(i=1; i<N; i++){
		temp = a[i];
		for(j=i; j >= 1; j--){
			if(a[j-1] > temp){
				a[j] = a[j-1];
			}else{
				break;
			}
		}
		a[j] = temp;

	}
}

int main(){
	int N;
	int *A;
	int **kind;
	int n_kind=0;
	int i,j;
	int duplicated_num=0;
	int sum=0;
	
	scanf("%d", &N);
	sum = N;
	
	A = (int*)malloc(sizeof(int) * N);
	for(i=0; i<N; i++){
		scanf("%d", A + i);
	}
	
	kind = (int**)malloc(sizeof(int*) * N);
	for(i=0; i<N; i++){
		kind[i] = (int*)malloc(sizeof(int) * 2);
	}

	//あとでの扱いを考えてソートする
	sort(A, N);
//	for(i=0; i<N; i++)
//		printf("%d ", A[i]);
	
	kind[n_kind][0] = A[0];
	kind[n_kind][1] = 1;

	for(i=1; i<N; i++){
		if(A[i] == kind[n_kind][0]){
			kind[n_kind][1]++;
		}else{
			n_kind++;
			kind[n_kind][1]=1;
			kind[n_kind][0]=A[i];
		}
	}	

//	for(i=0; i<=n_kind; i++){
//		printf("%d %d\n", kind[i][0], kind[i][1]);
//	}	


	for(i=0; i<=n_kind; i++){
		// ３枚以上あれば、それら３枚のうち２枚を食べる操作を繰り返す。
		if(kind[i][1] >= 3){
			kind[i][1] = (kind[i][1]+1) % 2 + 1;
			// その結果２枚減る。
			sum -= 2;
		}
	}
	//各カードは１枚か２枚になってる
//	for(i=0; i<=n_kind; i++){
//		printf("%d %d\n", kind[i][0], kind[i][1]);
//	}	
	for(i=0; i<=n_kind; i++){
		if(kind[i][1] == 2){
			duplicated_num++;
		}
	}
	
//	printf("sum = %d, duplicated = %d\n", sum, duplicated_num);

	if(duplicated_num % 2 == 0){
		sum -= duplicated_num;
	}else{
		sum -= duplicated_num+1;
	}

	printf("%d\n", sum);
	return 0;


}