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.*;

public class Main {
    static OutputWriter out;

    public static void main(String[] args) throws Exception {
        MyScanner in = new MyScanner(System.in);
        out = new OutputWriter(System.out);
        int t = in.nextInt();
        while (t>0){
            int n = in.nextInt();
            int m = in.nextInt();
            String s = in.next();
            int[]p = new int[n+1];
            for (int i = 0; i < m; i++) {
                int r = in.nextInt();
                p[0]++;
                p[r]--;
            }
            for (int i = 1; i < n+1; i++) {
                p[i]+=p[i-1];
            }
            int[]ans = new int[26];
            for (int i = 0; i < n; i++) {
                ans[s.charAt(i)-'a']+=p[i]+1;
            }
            for (int item:ans
                 ) {
                out.print(item+" ");
            }
            out.printLine();
            t--;
        }
        out.flush();
    }

    static void MergeSort(long[] a, long[] b, long[] c, int p, int r) {
        if (p < r) {
            int q = (r + p) / 2;
            MergeSort(a, b, c, p, q);
            MergeSort(a, b, c, q + 1, r);
            Merge(a, b, c, p, q, r);
        }
    }

    static void Merge(long[] a, long[] b, long[] c, int p, int q, int r) {
        int n1 = q - p + 1;
        int n2 = r - q;
        long[] R = new long[n1 + 1];
        long[] L = new long[n2 + 1];
        long[] R1 = new long[n1];
        long[] L1 = new long[n2];
        long[] R2 = new long[n1];
        long[] L2 = new long[n2];
        for (int i = 0; i < n1; i++) {
            R[i] = a[p + i];
            R1[i] = b[p + i];
            R2[i] = c[p + i];
        }
        R[n1] = Long.MAX_VALUE;
        for (int i = 0; i < n2; i++) {
            L[i] = a[q + i + 1];
            L1[i] = b[q + i + 1];
            L2[i] = c[q + i + 1];
        }
        L[n2] = Long.MAX_VALUE;
        int n = a.length;
        int j = 0;
        int k = 0;
        for (int i = p; i <= r; i++) {
            if (L[j] <= R[k]) {
                a[i] = L[j];
                b[i] = L1[j];
                c[i] = L2[j];
                j++;
            } else {
                a[i] = R[k];
                b[i] = R1[k];
                c[i] = R2[k];
                k++;
            }


        }
    }
}

class Graph {


    int n;
    ArrayList<Integer>[] adjList;

    public Graph(int n) {
        this.n = n;
        adjList = new ArrayList[n];
        for (int i = 0; i < n; i++)
            adjList[i] = new ArrayList<>();
    }

}

class pair implements Comparable {

    int key;
    int value;

    public pair(Object key, Object value) {

        this.key = (int) key;
        this.value = (int) value;

    }

    @Override
    public int compareTo(Object o) {
        pair temp = (pair) o;
        return key - temp.key;
    }
}

class MyScanner {

    StringTokenizer st;
    BufferedReader br;

    public MyScanner(InputStream s) {
        br = new BufferedReader(new InputStreamReader(s));
    }

    public MyScanner(FileReader f) {
        br = new BufferedReader(f);
    }

    public String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public String nextLine() throws IOException {
        return br.readLine();
    }

    public double nextDouble() throws IOException {
        String x = next();
        StringBuilder sb = new StringBuilder("0");
        double res = 0, f = 1;
        boolean dec = false, neg = false;
        int start = 0;
        if (x.charAt(0) == '-') {
            neg = true;
            start++;
        }
        for (int i = start; i < x.length(); i++)
            if (x.charAt(i) == '.') {
                res = Long.parseLong(sb.toString());
                sb = new StringBuilder("0");
                dec = true;
            } else {
                sb.append(x.charAt(i));
                if (dec)
                    f *= 10;
            }
        res += Long.parseLong(sb.toString()) / f;
        return res * (neg ? -1 : 1);
    }

    public boolean ready() throws IOException {
        return br.ready();
    }

    public boolean nextEmpty() throws IOException {
        String s = nextLine();
        st = new StringTokenizer(s);
        return s.isEmpty();
    }
}

class OutputWriter {
    private final PrintWriter writer;

    public OutputWriter(OutputStream outputStream) {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
    }

    public OutputWriter(Writer writer) {
        this.writer = new PrintWriter(writer);
    }

    public void print(Object... objects) {
        for (int i = 0; i < objects.length; i++) {
            if (i != 0)
                writer.print(' ');
            writer.print(objects[i]);
        }
    }


    public void printLine(Object... objects) {
        print(objects);
        writer.println();
    }

    public void close() {
        writer.close();
    }

    public void flush() {
        writer.flush();
    }

}