A. Magnets

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.

Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.

Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.

Output

On the single line of the output print the number of groups of magnets.

Sample test(s)

input

6
10
10
10
01
10
10

output

3

input

4
01
01
10
10

output

2

Note

The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.

The second testcase has two groups, each consisting of two magnets.

解题说明：此题是磁铁同性相斥异性相吸原理，不过由于题目中只有10,01两种数字，转换过来就是相同的数字相吸【10的0和10的1相吸】，不同的数字相斥，故只需要判断数字是否相同即可。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int main()
{
	int t,i,j,ans;
	scanf("%d",&t);
	scanf("%d",&i);
	t--;
	ans=1;
	while(t--)
	{
		scanf("%d",&j);
		if(j!=i)
		{
			ans++;
		}
		i=j;
	}
	printf("%d\n",ans);
	return 0;
}