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:
import java.io.*; 
import java.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
public class Main 
{ 
    
   // static int n;
    static HashSet<Integer> adj[];
    
     static boolean vis[];
   //  static long ans[];
   // static int arr[];
    static long mod=1000000007;
    static final long oo=(long)1e18;
    // static int n;
     
    public static void main(String[] args) throws IOException { 
    //    Scanner sc=new Scanner(System.in);
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        br = new BufferedReader(new InputStreamReader(System.in));
        int test=nextInt();
        //int test=1;
        outer:while(test-->0){
            int n=nextInt();
            int m=nextInt();
            String str=next();
            int arr[]=new int[m];
            for(int i=0;i<m;i++){
                arr[i]=nextInt()-1;
            }
            Arrays.sort(arr);
            int count[]=new int[26];
            int j=m;
            int k=0;
            for(int i=0;i<n;i++){
                int idx=str.charAt(i)-'a';
               // pw.println(m-k);
                count[idx]+=(m-k)+1;
                while(k<m&&arr[k]==i){
                    k++;
                }
            }
            for(int i=0;i<26;i++){
                pw.print(count[i]+" ");
            }
            pw.println();
        } 
        pw.close();
    }
    static class Pair implements Comparable<Pair>{
        int time;
        HashMap<Integer,ArrayList<Integer> > hm;
        int maxbooks;
        int i;
        int sum;
        Pair(int time,HashMap<Integer,ArrayList<Integer> > hm,int maxbooks, int i,int sum){
            this.time=time;
    
            this.hm=hm;
            this.maxbooks=maxbooks;
            this.i=i;
            this.sum=sum;
        }   
        public int compareTo(Pair p){
            if(time==p.time)
                return (sum-p.sum);
            return time-p.time;

        }
    }
    
    static void decToBinary(int n) 
    { 
        // array to store binary number 
        int[] binaryNum = new int[1000]; 
   
        // counter for binary array 
        int i = 0; 
        while (n > 0)  
        { 
            // storing remainder in binary array 
            binaryNum[i] = n % 2; 
            n = n / 2; 
            i++; 
        } 
   
        // printing binary array in reverse order 
        for (int j = i - 1; j >= 0; j--) 
            System.out.print(binaryNum[j]); 
        System.out.println();
    } 

    static long ncr(long n,long r){
        if(r==0)
            return 1;
        long val=ncr(n-1,r-1);
        val=(n*val)%mod;
        val=(val*modInverse(r,mod))%mod;
        return val;
    }

    static int find(ArrayList<Integer> a,int i){
        
        if(a.size()==0||i<0)return 0;
        ArrayList<Integer> l=new ArrayList<Integer>();
        ArrayList<Integer> r=new ArrayList<Integer>();
        for(int v:a){
            if((v&(1<<i))!=0)l.add(v);
            else r.add(v);
        }

        if(l.size()==0)return find(r,i-1);
        if(r.size()==0)return find(l,i-1);
        return Math.min(find(l,i-1),find(r,i-1))+(1<<i);
    }
    
    
    public static BufferedReader br;
    public static StringTokenizer st;
    public static String next() {
        while (st == null || !st.hasMoreTokens()) {
       try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
 
    public static Integer nextInt() {
        return Integer.parseInt(next());
    }
    public static Long nextLong() {
        return Long.parseLong(next());
    }
 
    public static Double nextDouble() {
        return Double.parseDouble(next());
    }
    // static class Pair{
    //     int x;int y;
    //     Pair(int x,int y,int z){
    //         this.x=x;
    //         this.y=y;
    //        // this.z=z;
    //      //   this.z=z;
    //       //  this.i=i;
    //     }
    // }
    // static class sorting implements Comparator<Pair>{
    //     public int compare(Pair a,Pair b){
    //         //return (a.y)-(b.y);
    //         if(a.y==b.y){
    //             return -1*(a.z-b.z);
    //         }
    //         return (a.y-b.y);
    //     }
   // }
    public static int[] na(int n)throws IOException{
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    
    // static class sorting implements Comparator<Pair>{  
    //     public int compare(Pair a1,Pair a2){  
    //         if(o1.a==o2.a)
    //             return (o1.b>o2.b)?1:-1;  
    //         else if(o1.a>o2.a)  
    //             return 1;  
    //         else  
    //             return -1;  
    //     }
    // }  
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static long gcd(long a, long b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    // To compute x^y under modulo m 
    static long power(long x, long y, long m){ 
        if (y == 0) 
            return 1;      
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    }
    static long fast_pow(long base,long n,long M){
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
       static long modInverse(long n,long M){
        return fast_pow(n,M-2,M);
    }
    // (1,1)   

    //     (3,2)   (3,5)

}