Case of the Zeros and Ones

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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 adjacent positions 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

41100

Output

0

Input

501010

Output

1

Input

811101111

Output

6

程序分析：此题的意思就是找出0比1多多少个，或者是1比0多多少个。也就是0和1的差。所以我需要分别统计出0和1的个数就好，然后在做差运算即可。

程序代码：

#include<iostream>

using namespace std;
char a[200005];
int main()
{
    int n,l,b=0,t,c=0;
    cin>>n;
    for(l=0;l<n;l++)
    {    cin>>a[l];

        if(a[l]==‘1‘)
        c++;
        else b++;
    }
        if(b>c)
        t=b-c;
        else t=c-b;
    cout<<t<<endl;
        return 0;
}