CF1272C

Recently, Norge found a string s=s1s2…sns=s1s2…sn consisting of nn lowercase Latin letters. As an exercise to improve his typing speed, he decided to type all substrings of the string ss. Yes, all n(n+1)2n(n+1)2 of them!

A substring of ss is a non-empty string x=s[a…b]=sasa+1…sbx=s[a…b]=sasa+1…sb (1≤a≤b≤n1≤a≤b≤n). For example, "auto" and "ton" are substrings of "automaton".

Shortly after the start of the exercise, Norge realized that his keyboard was broken, namely, he could use only kk Latin letters c1,c2,…,ckc1,c2,…,ck out of 2626.

After that, Norge became interested in how many substrings of the string ss he could still type using his broken keyboard. Help him to find this number.

Input

The first line contains two space-separated integers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 1≤k≤261≤k≤26) — the length of the string ss and the number of Latin letters still available on the keyboard.

The second line contains the string ss consisting of exactly nn lowercase Latin letters.

The third line contains kk space-separated distinct lowercase Latin letters c1,c2,…,ckc1,c2,…,ck — the letters still available on the keyboard.

Output

Print a single number — the number of substrings of ss that can be typed using only available letters c1,c2,…,ckc1,c2,…,ck.

Examples

input

7 2
abacaba
a b

output

12

input

10 3
sadfaasdda
f a d

output

21

input

7 1
aaaaaaa
b

output

0

Note

In the first example Norge can print substrings s[1…2]s[1…2], s[2…3]s[2…3], s[1…3]s[1…3], s[1…1]s[1…1], s[2…2]s[2…2], s[3…3]s[3…3], s[5…6]s[5…6], s[6…7]s[6…7], s[5…7]s[5…7], s[5…5]s[5…5], s[6…6]s[6…6], s[7…7]s[7…7].

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题倒是简单但是s=200000这样的数据用int就中招了

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#define LL long long
using namespace std;

int main()
{
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    set<char> all;
    for (int i = ; i < k; i++)
    {
        char t;
        cin >> t;
        all.insert(t);
    }
    LL cnt=,sum=;
    for (int i = ; i < s.size(); i++)
    {
        if (all.find(s[i]) != all.end())
        {
            cnt++;
        }
        else
        {
            sum += cnt*(cnt + ) / ;
            cnt = ;
        }
        if(i==s.size()- && all.find(s[i]) != all.end())
            sum += cnt*(cnt + ) / ;
    }
    cout << sum;
    //system("pause");
    return ;
}