Problem:
Score : 300 points

Problem Statement

You are playing the following game with Joisino.

Initially, you have a blank sheet of paper.

Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times.

Then, you are asked a question: How many numbers are written on the sheet now?

The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?

Constraints

1≤N≤100000

1≤A_i≤1000000000(=10^9)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print how many numbers will be written on the sheet at the end of the game.

Sample Input 1

3
6
2
6

Sample Output 1

1

The game proceeds as follows:

6 is not written on the sheet, so write 6.

2 is not written on the sheet, so write 2.

6 is written on the sheet, so erase 6.

Thus, the sheet contains only 2 in the end. The answer is 1.

Sample Input 2

4
2
5
5
2

Sample Output 2

0

It is possible that no number is written on the sheet in the end.

Sample Input 3

6
12
22
16
22
18
12

Sample Output 3

2

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        //最後の数値計算
        int answer = 0;

        //入力部
        Scanner sc = new Scanner(System.in);
        int roopTimes = Integer.parseInt(sc.nextLine());

        ArrayList<String> inputAllNumbers = new ArrayList<>();
        HashSet<String> checkNumbers = new HashSet<>();            //探索時に使うやつ、重複をなくすためHashSet

        for (int i = 0; i < roopTimes; i++) {
            String inputNumber = sc.nextLine();
            inputAllNumbers.add(inputNumber);
            checkNumbers.add(inputNumber);
        }


        //@TODO この書き方はまだ完全に理解しきれてないので要勉強
        //表記方法はメソッド参照(Java8からの書き方）、クラス名:メソッド名で呼び出す
        //あー、なんとなくわかった
        //これsortの方法としてStringのcompareToを呼び出してるってっていう、ものすごいストレートな書き方なんだな
        //いいなこれ、かっこいい


        inputAllNumbers.sort(String::compareTo);

        for (String key : checkNumbers) {
            if (countSearch(inputAllNumbers, key) % 2 != 0) {
                answer += 1;
            }
        }

        System.out.println(answer);
    }

    //線形探索メソッド
    //あんまり機能を詰め込むと汎用性がなくなるので、単純にカウントして回数を返すだけにする
    static int countSearch(ArrayList<String> checkTarget, String key) {
        int count = 0;

        for (int i = 0; i < checkTarget.size(); i++) {
            if (checkTarget.get(i).equals(key)) {
                count += 1;
            }

            if(count > 0 && !checkTarget.get(i).equals(key)){
                break;
            }
        }
        return count;
    }
}