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.util.*;
import java.lang.*;
public class Cf1272C {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		long l,p,i,j,q,s=0,t=-1;
		l=sc.nextLong();
		p=sc.nextLong();
		sc.nextLine();
		String str=sc.nextLine();
		char []brr=new char[(int)p];
		for (i=0;i<p;i++)
			brr[(int)i]=sc.next().charAt(0);	
		for(i=0;i<l;i++) {
			q=0;
			t++;
			for(j=0;j<p;j++) {
				if(str.charAt((int)i)==brr[(int)j]) {
					q=1;
					break;
				}
			}
			if(q==0) {
				s=s+(t*(t+1))/2;
				t=-1;
			}
			if(i==l-1 && q==1) {
				t++;
				s=s+(t*(t+1))/2;
			}
		}
			System.out.println(s);
		sc.close();
	}

}
