codeforces 873B Balanced Substring

You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string s_ls_l + 1s_l + 2... s_r, 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：给你一个01串，求出一个最长的子串，使得1的数量=0的数量

水题，把0变成-1，求前缀和

如果前缀和为0则有1～i

或有a[l]=a[r]则存在l+1~r

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
char s[100001];
int n,a[100001];
int Max[200005],Min[200005],ans;
int main()
{int i;
  cin>>n;
  scanf("%s",s);
  for (i=1;i<=n;i++)
    {
      if (s[i-1]=='1') a[i]=1;
      else a[i]=-1;
    }
  memset(Min,127,sizeof(Min));
  for (i=1;i<=n;i++)
    {
      a[i]+=a[i-1];
      if (a[i]==0) ans=max(ans,i);
      Max[a[i]+n]=max(i,Max[a[i]+n]);
      Min[a[i]+n]=min(i,Min[a[i]+n]);
      ans=max(ans,Max[a[i]+n]-Min[a[i]+n]);
    }
  cout<<ans;
}