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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <iomanip>
#include <cmath>
#include <vector>
#include <sstream>
#include <string>
#include <cstdio>
#include <stack>
#include <queue>
#include <list>
#include <numeric> //accumulate
#include <unordered_map> //ハッシュ関数

using namespace std;

//デバッグ時はTESTを定義する。本番時はこの行をコメントアウトする
//#define TEST //*******************************************************************************************************************************************

//そうすることで、以下に定義する、デバッグ時にのみ標準出力する、doutコマンドが実現できる。本番時はダミーストリームに出力
#ifdef TEST
    #define dout cout
#else
    stringstream dummy; //デバッグ時のdoutコマンドのダミー出力先
    #define dout dummy.str(""); dummy.clear(stringstream::goodbit); dummy //dummyを毎回リセットしておかないとメモリ溢れちゃう。
    //どうせ表示しないからgoodbitしなくていいけど、表示する場合はこれしておかないとバグるらしいので、勉強のために。詳しくは http://d.hatena.ne.jp/linden/20060427/p1
#endif


#define disp(A) dout << #A << " = " << (A) << endl
#define disP(A) dout << (A) << " "
#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)

#define dispAll(A,n) dout << #A << " = "; rep(j, 0, (n)) {disP(A[j]);} dout << endl

const int INF = 1e9;


int main(void) {
    int N;
    cin >> N;
    
    int a[N];
    rep(i,0,N) cin >> a[i];
    
    dispAll(a,N);
    
    sort(a,a+N);
    
    dispAll(a,N);
    
    int number;
    int cnt;
    int i = 0;
    int ans = 0;
    
    while(i<N) {
        
        
        number = a[i];
        cnt = 1;

        dout << "====\n";
        disp(i);
        disp(number);
        disp(cnt);
        
        i++;
        
        while( number == a[i] ) {
            dout << "---\n";
            disp(i);
            disp(a[i]);
            
            cnt++;
            i++;
            
            
            disp(cnt);
        }
        
        //奇数個のもののみカウント
        if( cnt%2 ) ans++;
        
    }
    

    cout << ans << endl;
    
    
    return 0;
}