CodeForces 556A Case of the Zeros and Ones

题目链接：http://codeforces.com/problemset/problem/556/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
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.

Examples

Input

4
1100

Output

0

Input

5
01010

Output

1

Input

8
11101111

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:


.

思路：STL的stack。先将最初的一个字符入栈，然后枚举整个串，如果栈非空且栈顶和当前字符是01串，就弹出栈顶字符，否则将字符入栈。最后只要检查栈内有几个字符即可。

附上AC代码：

#include <bits/stdc++.h>
using namespace std;
stack<char> s;
string str;
int n;

int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n >> str;
s.push(str[0]);
for (int i=1; i<str.size(); ++i){
if (str[i]=='0' && !s.empty() && s.top()=='1')
s.pop();
else if (str[i]=='1' && !s.empty() && s.top()=='0')
s.pop();
else
s.push(str[i]);
}
cout << s.size() << endl;
return 0;
}