Alice, Bob, Two Teams---CodeForces - 632B （前後綴）

                                          Alice, Bob, Two Teams

                                              Time limit   1500 ms    Memory limit  262144 kB

Alice and Bob are playing a game. The game involves splitting up game pieces into two teams. There are n pieces, and the i-th piece has a strength pi.

The way to split up game pieces is split into several steps:

1. First, Alice will split the pieces into two different groups A and B. This can be seen as writing the assignment of teams of a piece in an n character string, where each character is A or B.
2. Bob will then choose an arbitrary prefix or suffix of the string, and flip each character in that suffix (i.e. change A to B and B to A). He can do this step at most once.
3. Alice will get all the pieces marked A and Bob will get all the pieces marked B.

The strength of a player is then the sum of strengths of the pieces in the group.

Given Alice's initial split into two teams, help Bob determine an optimal strategy. Return the maximum strength he can achieve.

Input

The first line contains integer n (1 ≤ n ≤ 5·105) — the number of game pieces.

The second line contains n integers pi (1 ≤ pi ≤ 109) — the strength of the i-th piece.

The third line contains n characters A or B — the assignment of teams after the first step (after Alice's step).

Output

Print the only integer a — the maximum strength Bob can achieve.

Examples

Input

5
1 2 3 4 5
ABABA

Output

11

Input

5
1 2 3 4 5
AAAAA

Output

15

Input

1
1
B

Output

1

Note

In the first sample Bob should flip the suffix of length one.

In the second sample Bob should flip the prefix or the suffix (here it is the same) of length 5.

In the third sample Bob should do nothing.

簡單的理解就是 有一些卡片，每個卡片有相應的價值，卡片有兩面 A和B，Bob可以得到所有B面的卡片的價值

這是Bob可以對這些卡片進行一次翻轉，當然 只能翻轉前面的某一部分或者後面的某一部分（前綴或者後綴），問你Bob如何可以使得自己獲得的卡片的價值最大

我們使用兩個數組 sa[]和sb[]分別表示到第i張卡片，Alice和Bob可以獲得的價值

最重要的就是翻轉問題了，我使用了暴力的方法，詳情見代碼

#include<cstdio>
#include<string>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int inf=0x3f3f3f3f;
const int mm=5e5+5;

int n;
int a[mm];
ll sa[mm],sb[mm];
char str[mm];
ll sum;

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	scanf("%s",str+1);
	for(int i=1;i<=n;i++){
		sa[i]=sa[i-1];
		sb[i]=sb[i-1];
		if(str[i]=='A')
			sa[i]+=a[i];
		else 
			sb[i]+=a[i];	
	}
	ll res=max(sa[n],sb[n]);//初定最大值 不翻或者全翻 
	for(int i=1;i<=n;i++){
		res=max(res,sb[n]-sb[i]+sa[i]);//修改前綴  
		res=max(res,sa[n]-sa[i]+sb[i]);//修改後綴  
	} 
	cout<<res<<endl; 
		
	return 0;
}