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.
Examples
input
41100
output
0
input
501010
output
1
input
811101111
output
6
Note
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: 
.
题意：大致意思是给出一个01串，可删去01或10直到不能再删，输出最终01串的长度
思路：这是一道贪心，顺序什么的并不重要。因此本着贪到贪无可贪的原则，我们只需输出所有的0和1的个数再减一减就可以A掉了
代码：
var
s:array[1..200005] of char;
i,j,n,k,m1,m2:longint;
begin
readln(n);
for i:=1 to n do
begin
read(s[i]);
if s[i]=‘1‘ then inc(m1);
if s[i]=‘0‘ then inc(m2);
end;
writeln(abs(m1-m2));
end.