codeforces 519D dp

D. A and B and Interesting Substrings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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?

Input

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.

Output

Print the answer to the problem.

Examples

input

1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab

output

2

input

1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa

output

2

Note

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: 给定每个字母的价值, 求给定的字符串的子串中 首尾相同并且子串中不包含首尾的字符的价值和等于0的子串个数.

分析: 这种题一般就跟前缀和相关了, 一定要考虑前缀和啊, 有了前缀和的话, 当碰见一个字符的时候,假设现在前缀和是sum了(还没加上当前这个元素的价值), 考虑在他前面跟他相同的字符, 如果这个字符的前缀和也是sum(加上了这个字符的价值), 那么两者相减就是他们之间的元素的值的和, 也就是0. 那么, 只要记住相同元素的每个前缀和的个数, 在每次遍历的时候加上前面相同元素的这个前缀和的个数, 再考虑以下他的贡献, 就行了.

#include

#define inf 0x3f3f3f3f
using namespace std;

typedef long long ll;
typedef pair pii;

const int N=100010,MOD=1e9+7;
map dp[26];

int val[26];
string s;
int main()
{
    for(int i=0;i<26;i++)
    {
        cin>>val[i];
    }
    cin>>s;
    int n=s.size();
    ll sum=0,ans=0;
    for(int i=0;i