A. Case of the Zeros and Ones
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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: 

水题

直接计算s中0和1的个数

然后相减即可

 string s;
 int a,b;

 int main()
 {
     ios_base::sync_with_stdio();

     int n;

     cin>>n;
     cin>>s;

     ;i<n;i++)
     {
         ') a++;
         else b++;
     }

     cout<<abs(a-b)<<endl;

     ;
 }

#include<cmath>#include<math.h>#include<ctype.h>#include<algorithm>#include<bitset>#include<cassert>#include<cctype>#include<cerrno>#include<cfloat>#include<ciso646>#include<climits>#include<clocale>#include<complex>#include<csetjmp>#include<csignal>#include<cstdarg>#include<cstddef>#include<cstdio>#include<cstdlib>#include<cstring>#include<ctime>#include<cwchar>#include<cwctype>#include<deque>#include<exception>#include<fstream>#include<functional>#include<iomanip>#include<ios>#include<iosfwd>#include<iostream>#include<istream>#include<iterator>#include<limits>#include<list>#include<locale>#include<map>#include<memory>#include<new>#include<numeric>#include<ostream>#include<queue>#include<set>#include<sstream>#include<stack>#include<stdexcept>#include<streambuf>#include<string>#include<typeinfo>#include<utility>#include<valarray>#include<vector>#include<string.h>#include<stdlib.h>#include<stdio.h>using namespace std;#define havetxt 0#define shuru "input.txt"#define shuchu "output.txt"#define inf 0x7fffffff#define esp 1e-9#define pb push_back#define mp make_pairtypedef long long ll;
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----~.^^.'_!!/<}r[][?=r[[rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr][]][7[c[]yp%Zx$66666$$96666666666G66666Z%%9%6U9Vfx9ZUG996Gf6kxZPZZPPZP6k6PP6kkZZPPPPGPPyg(<<";)"**":>(/!-t0_0g,****""""})>43.  ~- ^_.'      -` `-' r9!`....^^'_\,^!"7r7rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr][[]r7[17rV6VG%$%%9999$$$$$$$$$$99999999VG$G%FEmFAwFZpkV%UxFL404T84484aYaxYxUsLsaaaaaaaay%}"*">>"""*)>;1T0kS)I4}"""*"****);Ti -`. ~1v -^ `~-   `` ~ _x]-.'^.''^''_("=rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr[[[[r7rii=[;"+;?}+++;;;;;;;;;;;;}}}}+++;;=?>181e%y0S&we^_!=U:`^.```--`._.`_-`^'--````---vL,*>*>}>"{"<)}[O5a[><[<"""*"***>)]U) ~-~~^5u `~  ``  `-``-~`uU^`,^~_!'!_)}r[[r[rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr[[[[[[[[[[[[[[[[[[[[rr[]?7=r111iiii]][][]]]]]]]][[]]]]]i1[c?nh'!}CV9xwfi)(}L8~~.`-.^--(70..^.-`,------~~)x_=">*>***::>)()>*_}>*\***"*"""*(vs`'``-.`a?^-`^`  `-`` ~-~~(y<`_'~_!_)";r]r7r][[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[*/
string s;int a,b;
int main(){    ios_base::sync_with_stdio(0);
int n;cin>>n;cin>>s;for(int i=0;i<n;i++){if(s[i]=='0') a++;else b++;}cout<<abs(a-b)<<endl;
return 0;}

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