题干：

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.

题目大意：

     A和B两个人玩游戏，给n个数，每个数对应一个 ' A ' 或 ' B ' ， 现在B玩家可以选一个前缀或者一个后缀，把A变B，B变A，操作结束后，B玩家可以取走所有标号为 ' B ' 的数字，问B玩家可以取到的数字之和最大是多少？ 

解题报告：

     看题目名称，猜是博弈？然而让你失望了，这题和A没有任何卵关系，，，先预处理出前缀A和B，后缀A和B，这四个数组。然后枚举每一个元素，维护翻这一位或者不翻这一位的最大值maxx，并且维护和ans的最大值，输出ans即可。

AC代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAX = 5e5 + 5; 
ll val[MAX];
ll a[MAX],b[MAX],A[MAX],B[MAX];
ll ans,maxx;
char s[MAX];
int main()
{int n;cin>>n;for(int i = 1; i<=n; i++) scanf("%lld",val+i);cin>>s+1;//prefix预处理 for(int i = 1; i<=n; i++) {if(s[i] == 'A') {a[i] = a[i-1] + val[i];b[i] = b[i-1];}else {a[i] = a[i-1];b[i] = b[i-1] + val[i];}}//suffix预处理for(int i = n; i>=1; i--) {if(s[i] == 'A') {A[i] = A[i+1] + val[i];B[i] = B[i+1];}else {A[i] = A[i+1];B[i] = B[i+1] + val[i];}} for(int i = 1; i<=n; i++) {maxx = max(b[i]+B[i+1],a[i] + B[i+1]);ans = max(ans,maxx);}for(int i = n; i>=1; i--) {maxx = max(b[i] + B[i+1],b[i] + A[i+1]);ans = max(ans,maxx);}printf("%lld\n",ans);return 0 ;
}