CodeForces 556A

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

41100

Output

0

Input

501010

Output

1

Input

811101111

Output

6

Hint

In the first sample test it is possible to change the string like the following: 1100→10→（empty）.

In the second sample test it is possible to change the string like the following: 01010→010→0.

In the third sample test it is possible to change the string like the following:11101111→111111.

题意:

给你一行整数，相邻两个数如果是01或者是10则去掉，如此下去，直到不可以操作为止，求出最后剩下的最短长度。

思路：

看1或0的个数，求出其最小的，用总长度减去这个最小长度的两倍。

代码：

#include<cstdio>#include<cstring>using namespace std;char  a[200010];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        scanf("%s",&a);        int a0=0,a1=0,ans=0;        for(int i=0;i<n;i++)        {            if(a[i]=='0')                a0++;        }        a1=n-a0;        if(a0<=a1)        ans=n-2*a0;        else   if(a1<a0)            ans=n-2*a1;        printf("%d\n",ans);    }    return 0;}