Problem: You want to perform the combo on your opponent in one popular fighting game. The combo is the string s consisting of n lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in s. I.e. if s="abca" then you have to press 'a', then 'b', 'c' and 'a' again.

You know that you will spend m wrong tries to perform the combo and during the i-th try you will make a mistake right after p_i-th button (1 ≤ p_i < n) (i.e. you will press first p_i buttons right and start performing the combo from the beginning). It is guaranteed that during the m+1-th try you press all buttons right and finally perform the combo.

I.e. if s="abca", m=2 and p = [1, 3] then the sequence of pressed buttons will be 'a' (here you're making a mistake and start performing the combo from the beginning), 'a', 'b', 'c', (here you're making a mistake and start performing the combo from the beginning), 'a' (note that at this point you will not perform the combo because of the mistake), 'b', 'c', 'a'.

Your task is to calculate for each button (letter) the number of times you'll press it.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Then t test cases follow.

The first line of each test case contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 2 ⋅ 10^5) — the length of s and the number of tries correspondingly.

The second line of each test case contains the string s consisting of n lowercase Latin letters.

The third line of each test case contains m integers p_1, p_2, ..., p_m (1 ≤ p_i < n) — the number of characters pressed right during the i-th try.

It is guaranteed that the sum of n and the sum of m both does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5, ∑ m ≤ 2 ⋅ 10^5).

It is guaranteed that the answer for each letter does not exceed 2 ⋅ 10^9.

Output

For each test case, print the answer — 26 integers: the number of times you press the button 'a', the number of times you press the button 'b', ..., the number of times you press the button 'z'.

Example

Input


3
4 2
abca
1 3
10 5
codeforces
2 8 3 2 9
26 10
qwertyuioplkjhgfdsazxcvbnm
20 10 1 2 3 5 10 5 9 4


Output


4 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 9 4 5 3 0 0 0 0 0 0 0 0 9 0 0 3 1 0 0 0 0 0 0 0 
2 1 1 2 9 2 2 2 5 2 2 2 1 1 5 4 11 8 2 7 5 1 10 1 5 2 

Note

The first test case is described in the problem statement. Wrong tries are "a", "abc" and the final try is "abca". The number of times you press 'a' is 4, 'b' is 2 and 'c' is 2.

In the second test case, there are five wrong tries: "co", "codeforc", "cod", "co", "codeforce" and the final try is "codeforces". The number of times you press 'c' is 9, 'd' is 4, 'e' is 5, 'f' is 3, 'o' is 9, 'r' is 3 and 's' is 1.

Solution (in JAVA):

Solution:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you

Sometime i think i am goona die and I want to hang myself cause i am not that good in cp ! :(
*/ 
  
import java.util.*;
import java.lang.Math; 
import java.util.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
     
    public class Main {
         static int oo = (int)1e9;
      public static void main(String[] args) {
    	       InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            Scanner in = new Scanner(inputStream);
            PrintWriter out = new PrintWriter(outputStream);
            MuskA solver = new MuskA();
            solver.solve(1, in, out);
            out.close();
    	}
    	
    static class MuskA {
       static int inf = (int) 1e9 + 7;
       public void solve(int testNumber, Scanner in, PrintWriter out) {
        StringBuilder output = new StringBuilder();
        int T = in.nextInt();
        for (int t=0; t<T; t++) {
            int N = in.nextInt();
            int M = in.nextInt();
            String S = in.next();
            int[][] stats = new int[S.length()+1][26];
            for (int n=1; n<=N; n++) {
                char c = S.charAt(n-1);
                for (int i=0; i<26; i++) {
                    stats[n][i] = stats[n-1][i];
                }
                stats[n][c-'a']++;
            }
            int[] answer = new int[26];
            for (int m=0; m<=M; m++) {
                int p = (m < M) ? in.nextInt() : N;
                for (int i=0; i<26; i++) {
                    answer[i] += stats[p][i];
                }
            }
            for (int i=0; i<26; i++) {
                output.append(answer[i]).append(' ');
            }
            output.append('\n');
        }
        out.print(output);
       }      
    	}     
    public int factorial(int n) {
            int fact = 1;
            int i = 1;
           while(i <= n) {
             fact *= i;
             i++;
           }
            return fact;
    }

    public static long gcd(long x,long y)
    {
    	if(x%y==0)
    		return y;
    	else
    		return gcd(y,x%y);
    }
    public static int gcd(int x,int y)
    {
    	if(x%y==0)
    		return y;
    	else 
    		return gcd(y,x%y);
    }
    public static int abs(int a,int b)
    {
    	return (int)Math.abs(a-b);
    }
    public static long abs(long a,long b)
    {
    	return (long)Math.abs(a-b);
    }
    public static int max(int a,int b)
    {
    	if(a>b)
    		return a;
    	else
    		return b;
    }
    public static int min(int a,int b)
    {
    	if(a>b)
    		return b;
    	else 
    		return a;
    }
    public static long max(long a,long b)
    {
    	if(a>b)
    		return a;
    	else
    		return b;
    }
    public static long min(long a,long b)
    {
    	if(a>b)
    		return b;
    	else 
    		return a;
    }
     
     
    public static long pow(long n,long p,long m)
    {
    	 long  result = 1;
    	  if(p==0)
    	    return 1;
    	if (p==1)
    	    return n;
    	while(p!=0)
    	{
    	    if(p%2==1)
    	        result *= n;
    	    if(result>=m)
    	    result%=m;
    	    p >>=1;
    	    n*=n;
    	    if(n>=m)
    	    n%=m;
    	}
    	return result;
    }
    public static long pow(long n,long p)
    {
    	long  result = 1;
    	  if(p==0)
    	    return 1;
    	if (p==1)
    	    return n;
    	while(p!=0)
    	{
    	    if(p%2==1)
    	        result *= n;	    
    	    p >>=1;
    	    n*=n;	    
    	}
    	return result;
     
    }
     
    static long sort(int a[]){  
         int n=a.length;
    	int b[]=new int[n];	
    	return mergeSort(a,b,0,n-1);
         
    }
     
    static long mergeSort(int a[],int b[],long left,long right){
         long c=0;
         if(left<right){   
          long mid=left+(right-left)/2;
    	 c= mergeSort(a,b,left,mid);
    	 c+=mergeSort(a,b,mid+1,right);
    	 c+=merge(a,b,left,mid+1,right); 
         }	
    	return c;	 
    }
    static long merge(int a[],int  b[],long left,long mid,long right){
         long c=0;int i=(int)left;int j=(int)mid; int k=(int)left;
         while(i<=(int)mid-1&&j<=(int)right){ 
              if(a[i]<=a[j]){
                   b[k++]=a[i++]; 
              }
              else{ 
                   b[k++]=a[j++];c+=mid-i;
              }
         }
         
         while (i <= (int)mid - 1)   
                 b[k++] = a[i++]; 
         while (j <= (int)right)
                 b[k++] = a[j++];
         for (i=(int)left; i <= (int)right; i++) 
    	        a[i] = b[i];
    	        return c; 
         }
     
  }