题目链接：http://codeforces.com/problemset/problem/873/B

ou are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.

You have to determine the length of the longest balanced substring of s.

Input

The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.

The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.

Output

If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.

Examples

input

8
11010111

output

4

input

3
111

output

0

Note

In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.

In the second example it's impossible to find a non-empty balanced substring.

题目大意：给出一个01串，找出连续的一个区间内0的个数和1的个数相等的最长子串

暴力超时，我们1看成+1，0看成-1，如果一个数字第一次出现，记录下来，如果再次出现，那么他们的0，1个数就相等，找出最大的那个数即可：

#include<stdio.h>
#include<string.h>  
#include<math.h>  
  
#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<iostream>  
#include<algorithm>  
using namespace std;  

#define ll long long  
#define INF 0x3f3f3f3f  
#define mod 1000000007
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b) 
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
//std::ios::sync_with_stdio(false);
const int N=100000;


char s[100100];
int sum[100100],num[200100];
int n;

int main()
{
	std::ios::sync_with_stdio(false);
	cin>>n;
	cin>>s;
	clean(sum,0);
	clean(num,INF);//初始化INF 
	int len=strlen(s);
	sum[0]=0;
	num[N]=0;
	int maxl=0;
	for(int i=1;i<=len;++i)
	{
		if(s[i-1]=='1')
			sum[i]=sum[i-1]+1;
		else
			sum[i]=sum[i-1]-1;
		if(num[N+sum[i]]==INF)//没出现过 
			num[N+sum[i]]=i;//记录初始位置 
		else
			maxl=max(maxl,i-num[N+sum[i]]);//出现过，目前位置为最长的长度 
	}
	cout<<maxl<<endl;
}