After an uphill battle, General Li won a great victory. Now the head of state decide to reward him with honor and treasures for his great exploit.

One of these treasures is a necklace made up of 26 different kinds of gemstones, and the length of the necklace is n. (That is to say: n gemstones are stringed together to constitute this necklace, and each of these gemstones belongs to only one of the 26 kinds.)

In accordance with the classical view, a necklace is valuable if and only if it is a palindrome - the necklace looks the same in either direction. However, the necklace we mentioned above may not a palindrome at the beginning. So the head of state decide to
cut the necklace into two part, and then give both of them to General Li.

All gemstones of the same kind has the same value (may be positive or negative because of their quality - some kinds are beautiful while some others may looks just like normal stones). A necklace that is palindrom has value equal to the sum of its gemstones'
value. while a necklace that is not palindrom has value zero.

Now the problem is: how to cut the given necklace so that the sum of the two necklaces's value is greatest. Output this value.

The first line of input is a single integer T (1 ≤ T ≤ 10) - the number of test cases. The description of these test cases follows.

For each test case, the first line is 26 integers: v₁, v₂, ..., v₂₆ (-100 ≤ v_i ≤ 100, 1 ≤ i ≤ 26), represent the value of gemstones of each kind.

The second line of each test case is a string made up of charactor 'a' to 'z'. representing the necklace. Different charactor representing different kinds of gemstones, and the value of 'a' is v₁, the value of 'b' is v₂, ..., and so on.
The length of the string is no more than 500000.

Output a single Integer: the maximum value General Li can get from the necklace.

2
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
aba
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
acacac

1
6

lcy

 

本题给定字母的权值和一个字符序列，若为回文串，则字符串的价值为该字符串的所有字符权值和；否则为0。现在要把字符串分成两部分，求两部分的权值和。

         对于某段字符串的权值和我们可以用累加和进行预处理。关键在于使左右两边的权值和最大。我们可以预处理求出左右两边的权值和，即求出左右两边是否为回文串，然后挑选其中最大的。

关于求回文串，可以用扩展KMP算法或用Manacher算法，我首先用的Manacher算法，但在求价值时出问题了；于是改用扩展KMP算法。

         首先处理从左端点开始的回文串，我们可以通过现在原串末尾加一个未曾出现的字符将其隔开，然后将原串反过来加在新串后面，这样就可以通过KMP算法的next[]数组来求从左端点开始的长度不一的回文串了；然后是求从右边端点开始的回文串，将原字符翻过来形成新串，再重复上面的操作。

 

/*
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

//*****************************************************************
const int MAXN=500000+100;
char str1[MAXN*2],str2[MAXN*2];//待处理字符串
int num[MAXN*2];
int value[30];
int sum[MAXN];
int pre[MAXN];
int suf[MAXN];


//将str1变成str2,如abab变成$#a#b#a#b#
void init()
{
	int i,id;
	str2[0]='$';sum[0]=0;
	str2[1]='#';sum[1]=0;
	for(i=0,id=2;str1[i];i++,id+=2)
	{
		str2[id]=str1[i];sum[id]=sum[id-1]+value[str1[i]-'a'];
		str2[id+1]='#';sum[id+1]=sum[id];
	}
	str2[id]=0;sum[id]=sum[id-1];
}
//Manacher算法求最长回文子串,时间复杂度为O(N)
void Manacher()
{
	int i,ans=0,MxR=0,pos,len=strlen(str2);
	memset(pre,0,sizeof(pre));memset(suf,0,sizeof(suf));
	for(i=1;str2[i];i++)
	{
		if(MxR>i)num[i]=num[pos*2-i]<(MxR-i)?num[pos*2-i]:(MxR-i);
		else num[i]=1;
		while(str2[i+num[i]]==str2[i-num[i]])
			num[i]++;
		if(num[i]+i>MxR)
		{
			MxR=num[i]+i;
			pos=i;
		}
		 if(i-num[i]==0)pre[i+num[i]-1]=1;//表示前缀(前p[i]-1个字符)是回文串   
          if(i+num[i]==len)suf[i-num[i]+1]=1;//表示后缀(后p[i]-1个字符)是回文串   

	}
}
//*****************************************************************


int main()
{
	int cas,i,len,tmp,ans;
	cin>>cas;
	while(cas--)
	{
		for(i=0;i<26;i++)
			scanf("%d",&value[i]);
		scanf("%s",str1);
		init();
		Manacher();

		len=strlen(str2);

		for(i=1;i<len;i++)
			printf("%d  ",pre[i]);
		printf("\n");
		for(i=1;i<len;i++)
			printf("%d  ",suf[i]);
		printf("\n");

		ans=1;
		for(i=1;i<len;i++)
		{
			tmp=0;

			if(pre[i]&&i+num[i]<len)tmp+=sum[i-1];
			if(suf[i]) tmp+=sum[i+num[i]]-sum[i-1];
			if(ans<tmp)ans=tmp;
		}
		printf("%d\n",ans);

	}
	return 0;
}
*/

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int MAXN=(500000+100)*2;
int value[MAXN],sum[MAXN],pre[MAXN],suf[MAXN];
char str[MAXN];

//*****************************************************************
int next[MAXN];
//KMP算法中计算next[]数组
void getNext(char *p)
{
    int j,k,len=strlen(p);
    j=0;
    k=-1;
    next[0]=-1;
    while(j<len)
    {
        if(k==-1||p[j]==p[k])
        {
            next[++j]=++k;
        }
        else k=next[k];
    }
}
//*****************************************************************
int main()
{
	int cas,i,len,tlen,ans,k,tmp;
	char ch;
	//freopen("in.txt","r",stdin);
	cin>>cas;
	while(cas--)
	{
		for(i=0;i<26;i++)
			scanf("%d",&value[i]);
		scanf("%s",str);

		len=strlen(str);
		sum[0]=value[str[0]-'a'];
		for(i=1;i<len;i++)
			sum[i]=sum[i-1]+value[str[i]-'a'];
		//sum[i]=sum[i-1];

		tlen=len*2+1;
		str[len]='#';
		for(i=0;i<len;i++)
		{
			str[len+1+i]=str[len-1-i];
		}
		str[len+len+1]=0;
	//	printf("str=%s\n",str);
		getNext(str);
		k=next[tlen];
		while(k!=0)
		{
			pre[k]=cas+2;
			k=next[k];
		}
		
		tlen=len*2+1;
		for(i=0;i<len/2;i++)
		{
			ch=str[i];
			str[i]=str[len-i-1];
			str[len-i-1]=ch;
		}
		for(i=0;i<len;i++)
		{
			str[len+1+i]=str[len-1-i];
		}	
		getNext(str);
		k=next[tlen];
		while(k!=0)
		{
			suf[k]=cas+2;
			k=next[k];
		}

		int ans=-99999999;
		for(i=1;i<len;i++)
		{
			tmp=0;
			if(pre[i]==cas+2)tmp+=sum[i-1];
			if(suf[len-i]==cas+2)tmp+=(sum[len-1]-sum[i-1]);
			if(ans<tmp)ans=tmp;
		}
		printf("%d\n",ans);
	}
	return 0;
}