Recently, Norge found a string s = s_1 s_2 … s_n consisting of n lowercase Latin letters. As an exercise to improve his typing speed, he decided to type all substrings of the string s. Yes, all (n (n + 1))/(2) of them!

A substring of s is a non-empty string x = s[a … b] = s_{a} s_{a + 1} … s_{b} (1 ≤ 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 k Latin letters c_1, c_2, …, c_k out of 26.

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

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 26) — the length of the string s and the number of Latin letters still available on the keyboard.

The second line contains the string s consisting of exactly n lowercase Latin letters.

The third line contains k space-separated distinct lowercase Latin letters c_1, c_2, …, c_k — the letters still available on the keyboard.

Output

Print a single number — the number of substrings of s that can be typed using only available letters c_1, c_2, …, c_k.

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[2…3], s[1…3], s[1…1], s[2…2], s[3…3], s[5…6], s[6…7], s[5…7], s[5…5], s[6…6], s[7…7].
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Makisar
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CYetAnotherBrokenKeyboard solver = new CYetAnotherBrokenKeyboard();
        solver.solve(1, in, out);
        out.close();
    }

    static class CYetAnotherBrokenKeyboard {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
        /*
        find all the ranges and then apply the n(n+1)/2 formula

        abacaba ab

        start = 0
        fori if c not in keyboard

        substring = i - start;

        aba aba = 6 + 6 = 12
         */
            int n = in.nextInt();
            int k = in.nextInt();
            char[] line = in.next().toCharArray();
            Set<Character> keyboard = new HashSet<>();
            for (int i = 0; i < k; i++) {
                keyboard.add(in.next().charAt(0));
            }

            long sum = 0L;
            for (int i = 0; i < n; i++) {
                int end = i;
                while (end < n && keyboard.contains(line[end])) end++;
                sum += (end - i) * 1L * (end - i + 1) / 2;
                i = end;
            }

            out.println(sum);
        }

    }
}

