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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#define MAX 100000

void sort(int A[], int N)
{
        int i, j, tmp;
        for(i=0; i<N; i++)
        {
                for(j=i; j<N; j++)
                {
                        if(A[i] > A[j])
                        {
                                tmp = A[i];
                                A[i] = A[j];
                                A[j] = tmp;
                        }
                }
        }
}

int main()
{
        int A[MAX];
        int N;
        int i, j;
        int cnt;

        scanf("%d", &N);
        for(i=0; i<N; i++)scanf("%d", &A[i]);

        sort(A, N);

        //for(i=0; i<N; i++)printf("%d ", A[i]);

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

        printf("%d\n", cnt);

        return 0;
}