ACM比较1和0的个数

Description

Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.

Once he thought about a string of length n consisting of zeroes and ones. Consider the following operation: we choose any two adjacentpositions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length n - 2 as a result.

Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.

Input

First line of the input contains a single integer n (1 ≤ n ≤ 2·10⁵), the length of the string that Andreid has.

The second line contains the string of length n consisting only from zeros and ones.

Output

Output the minimum length of the string that may remain after applying the described operations several times.

Sample Input

Input

4
1100

Output

0

Input

5
01010

Output

1

Input

8
11101111

Output

6

Hint

In the first sample test it is possible to change the string like the following: .

In the second sample test it is possible to change the string like the following: .

In the third sample test it is possible to change the string like the following: .

解题思路：

这个题目的大意就是给定一串只包含0和1的数字，当两个相邻位置的数分别是0和1时，我们就可以同时将这两个数歌去除掉，问最后还剩多少个数字。（其实也就是让我们判断0和1相差多少个）所以我们就只要判断一下0和1相差多少个就可以了。

程序代码：

#include <iostream>
using namespace std;
char a[200005];
int main()
{
    int n,x=0,y=0,v;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i];
    for(int k=0;k<n;k++)
    {
        if(a[k]=='0')
            x++;
        else 
            y++;
    }
    if(x>y)
        v=x-y;
    else 
        v=y-x;
    cout<<v<<endl;
    return 0;
}