DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter c DZY knows its value wc. For each special string s?=?s1s2... s|s| (|s| is the length of the string) he represents its value with a function f(s), where

Now DZY has a string s. He wants to insert k lowercase letters into this string in Z??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcmRlciB0byBnZXQgdGhlIGxhcmdlc3QgcG9zc2libGUgdmFsdWUgb2YgdGhlIHJlc3VsdGluZyBzdHJpbmcuIENhbiB5b3UgaGVscCBoaW0gY2FsY3VsYXRlIHRoZSBsYXJnZXN0IHBvc3NpYmxlIHZhbHVlIGhlIGNvdWxkIGdldD88L3A+CgoKCklucHV0CjxwPgpUaGUgZmlyc3QgbGluZSBjb250YWlucyBhIHNpbmdsZSBzdHJpbmcgPGVtPnM8L2VtPiAoMT+h3D8="s|?≤?103).

The second line contains a single integer k (0?≤?k?≤?103).

The third line contains twenty-six integers from wa to wz. Each such number is non-negative and doesn't exceed 1000.

Print a single integer ― the largest possible value of the resulting string DZY could get.

abc
3
1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

41

In the test sample DZY can obtain "abcbbc", value?=?1?1?+?2?2?+?3?2?+?4?2?+?5?2?+?6?2?=?41.

#include
  
   
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            using namespace std; #define For(i,n) for(int i=1;i<=n;i++) #define Fork(i,k,n) for(int i=k;i<=n;i++) #define Rep(i,n) for(int i=0;i
           
            =0;i--) #define Forp(x) for(int p=pre[x];p;p=next[p]) #define Lson (x<<1) #define Rson ((x<<1)+1) #define MEM(a) memset(a,0,sizeof(a)); #define MEMI(a) memset(a,127,sizeof(a)); #define MEMi(a) memset(a,128,sizeof(a)); #define INF (2139062143) #define F (100000007) #define MAXN (1000+10) long long mul(long long a,long long b){return (a*b)%F;} long long add(long long a,long long b){return (a+b)%F;} long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;} typedef long long ll; char s[MAXN]; int k,w[MAXN]; int main() { // freopen("Strings.in","r",stdin); // freopen(".out","w",stdout); scanf("%s\n%d",s+1,&k); int len=strlen(s+1); int p=0; Fork(i,'a','z') cin>>w[i],p=max(p,w[i]); ll ans=0; For(i,len) ans+=w[s[i]]*i; if (k) ans+=(len+1+len+k)*k/2*p; cout<