A and B are preparing themselves for programming contests.

After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.

A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).

B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).

Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.

Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?

The first line contains 26 integers x_a, x_b, ..., x_z ( - 10⁵ ≤ x_i ≤ 10⁵) — the value assigned to letters a, b, c, ..., z respectively.

The second line contains string s of length between 1 and 10⁵ characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.

Print the answer to the problem.

In the first sample test strings satisfying the condition above are abca and bcab.

In the second sample test strings satisfying the condition above are two occurences of aa.

题意：每一个字符都有一个权值，让你用Ｏ（n）的算法找到满足以下条件的子串个数

１.首位和末尾字母相同

２.除去首位和末尾的权值等于０

题解：我们只要找到该位置之前有多少个和这个位置有相同字母，且权值相同的位置就好

用一个map<ll,ll>就好，表示到第i个字母的时候，取得的和为j的个数有多少个

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 100001

#define eps 1e-9

const int inf=0x7fffffff;   //无限大

int main()

{

        ll ans=;

        map<ll,ll> kiss[];

        int point[];

        for(int i=;i<;i++)

        {

                cin>>point[i];

        }

        string s;

        cin>>s;

        ll sum=;

        for(int i=;i<s.size();i++)

        {

                ans+=kiss[s[i]-'a'][sum];

                sum+=point[s[i]-'a'];

                kiss[s[i]-'a'][sum]++;

        }

        cout<<ans<<endl;

}