【52.49%】【codeforces 556A】Case of the Zeros and Ones

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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: .

【题目链接】:http://codeforces.com/contest/556/problem/A

【题解】



让你消除相邻的01子串;

因为看到挺多人过的，就用函数写了个暴力;

然后就过了。

不过…998ms…….

用数组模拟链表应该更快.

不过挺麻烦的吧.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
#define pri(x) printf("%d",x)
#define prl(x) printf("%I64d",x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 2e5+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

string s;
int n;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin>>n;
    cin >> s;
    s=' '+s;
    int len = s.size();
    len--;
    int i = 1;
    while (i <= len-1)
    {
        if (s[i]!=s[i+1])
        {
            s.erase(i,2);
            i--;
            if (i==0) i=1;
        }
        else
            i++;
        len = s.size()-1;
    }
    cout << len;
    return 0;
}