CodeForces 632B Alice, Bob, Two Teams

B. Alice, Bob, Two Teams

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

這個題充分的考驗了一個人的英語水準。反正我看了很久纔看懂，沒愛了

主要是第二步，它用了個不常見的詞，結果就死活看不懂了……

第二步的大意是Bob可以選擇任意的前綴或者後綴，來翻轉他們（翻轉就是把A變成B，把B變成A）。

重點就在這個前綴或者後綴這！！！題目給的Note還死活看不出來什麼意思……全靠自己領悟。領悟出來就是個水題啊TAT

舉個例子吧，比如：

5

1 2 3 4 5

BBABB

因爲只能翻轉任意長度的前綴或者後綴，所以想要把A翻成B，我們只能翻（1 2 3）BBA或者（3 4 5）ABB。翻BBA結果是12，翻ABB結果是6，不翻轉結果也是12，所以最後輸出12。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
__int64 a[500000+10];
char ch[500000+10];
int main()
{
	int i,n;
	while(~scanf("%d",&n))
	{
		__int64 sum=0; 
		for(i=0;i<n;i++)
			scanf("%I64d",&a[i]);
		scanf("%s",ch);
		for(i=0;i<n;i++)
			if(ch[i]=='B')
				sum+=a[i];
		__int64 ans=0,cnt=0;
		for(i=0;i<n;i++)
		{
			if(ch[i]=='A')
				ans+=a[i];
			else if(ch[i]=='B')
				ans-=a[i];
			cnt=max(ans,cnt);
		}
		ans=0;
		for(i=n-1;i>=0;i--)
		{
			if(ch[i]=='A')
				ans+=a[i];
			else if(ch[i]=='B')
				ans-=a[i];
			cnt=max(ans,cnt);
		}
		printf("%I64d\n",cnt+sum);
	}
	return 0;
}