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: Accepted)

#include <stdio.h>
#include <malloc.h>
typedef struct node
{
	int ai;
	int time;
	struct node*left;
	struct node*right;
}tree;
tree* create(int now)
{
	tree*r=(tree*)malloc(sizeof(tree));
	r->ai=now;
	r->time=1;
	r->left=NULL;
	r->right=NULL;
	return r;
}
void add(tree*root,int now)
{
	if(root->ai==now)
	{
		root->time++;
		return;
	}
	if(root->ai<now)
	{
		if(root->right!=NULL)
		{
			add(root->right,now);
			return;
		}
		else
		{
			root->right=create(now);
			return;
		}
	}
	else
	{
		if(root->left!=NULL)
		{
			add(root->left,now);
			return;
		}
		else
		{
			root->left=create(now);
			return;
		}
	}
}
int allxian=0;
void searchall(tree*root)
{
	if(root==NULL)
		return;
	searchall(root->left);
	searchall(root->right);
	if(root->time%2!=0)
	{
		allxian++;
	}
}
int main()
{
	int alla[100001];
	int allcount=0;
	int i,j;
	int now;
	tree*root;
	scanf("%d",&allcount);
	scanf("%d",&now);
	root=create(now);
	for(i=0;i<allcount-1;i++)
	{
		scanf("%d",&now);
		add(root,now);
	}
	searchall(root);
	printf("%d",allxian);
	return 0;
}