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:
//HashMap sortByValue(map);
//long comb=nCr(5,2);
//int gcd(s,l);
//Pair p=new Pair(x,y);

import java.util.*;
import java.io.*;

public class AirConditioner {

    InputStream is;
    PrintWriter out;
    String INPUT = "";

    void solve() {
        int t = ni();
        for (int i = 0; i < t; i++) {
            int n = ni();
            int m = ni();
            char[] s = ns(n);
            int p[] = new int[m];
            for (int j = 0; j < m; j++) {
                p[j] = ni() - 1;
            }
            
            
            Map<Integer,Integer> map=new HashMap<Integer,Integer>();
            for (int j = 0; j < m; j++) {
                map.put(p[j], map.getOrDefault(p[j], 0)+1);
            }
            
            List<Integer> al=new ArrayList<>(map.keySet());
            Collections.sort(al);
            int x = m + 1;
            int k = 0;
            int[] a = new int[26];
            for (int j = 0; j < n; j++) {
                a[s[j]-97] = a[s[j]-97]+x;
                if (k<map.size() && j== al.get(k)) {
                    //a[(int) s[j] - 97][j] = x;
                    
                    x-=map.get(al.get(k));
                    k++;
                }

            }
            
            for (int j = 0; j < 26; j++) {
                out.print(a[j]+" ");
            }
            out.println();
        }

    }

    long nCr(int n, int k) {
        long C[] = new long[k + 1];
        // nC0 is 1 
        C[0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = Math.min(i, k); j > 0; j--) {
                C[j] = C[j] + C[j - 1];
            }
        }
        return C[k];
    }

    public int gcd(int s, int l) {
        if (s == 0) {
            return l;
        }

        return gcd(l % s, s);
    }

    public HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm) {
        // Create a list from elements of HashMap 
        List<Map.Entry<Integer, Integer>> list
                = new LinkedList<>(hm.entrySet());

        //->change o1,o2 for reverseorder
        Collections.sort(list, (Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) -> (o1.getValue()).compareTo(o2.getValue()));

        // put data from sorted list to hashmap  
        HashMap<Integer, Integer> temp = new LinkedHashMap<>();
        list.forEach((aa) -> {
            temp.put(aa.getKey(), aa.getValue());
        });
        return temp;
    }

    void run() throws Exception {
        is = System.in;//oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        out = new PrintWriter(System.out);

        long s = System.currentTimeMillis();
        solve();
        out.flush();
        tr(System.currentTimeMillis() - s + "ms");
    }

    public static void main(String[] args) throws Exception {
        new AirConditioner().run();
    }

    private byte[] inbuf = new byte[1024];
    public int lenbuf = 0, ptrbuf = 0;

    private int readByte() {
        if (lenbuf == -1) {
            throw new InputMismatchException();
        }
        if (ptrbuf >= lenbuf) {
            ptrbuf = 0;
            try {
                lenbuf = is.read(inbuf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (lenbuf <= 0) {
                return -1;
            }
        }
        return inbuf[ptrbuf++];
    }

    private boolean isSpaceChar(int c) {
        return !(c >= 33 && c <= 126);
    }

    private int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b));
        return b;
    }

    private double nd() {
        return Double.parseDouble(ns());
    }

    private char nc() {
        return (char) skip();
    }

    private String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    private char[] ns(int n) {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while (p < n && !(isSpaceChar(b))) {
            buf[p++] = (char) b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }

    private char[][] nm(int n, int m) {
        char[][] map = new char[n][];
        for (int i = 0; i < n; i++) {
            map[i] = ns(m);
        }
        return map;
    }

    private int[] na(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = ni();
        }
        return a;
    }

    private int ni() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }

        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    private long nl() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }

        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    private boolean oj = System.getProperty("ONLINE_JUDGE") != null;

    private void tr(Object... o) {
        if (!oj) {
            System.out.println(Arrays.deepToString(o));
        }
    }

    /*void swap(int l, int r, int[] a) {
        int temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }

    int order_statistics(int[] a, int n, int k) {
        k--;
        for (int l = 0, r = n - 1;;) {
            if (r <= l + 1) {
                // the current part size is either 1 or 2, so it is easy to find the answer
                if (r == l + 1 && a[r] < a[l]) {
                    swap(l, r, a);
                }
                return a[k];
            }

            // ordering a[l], a[l+1], a[r]
            int mid = (l + r) >> 1;
            swap(mid, l + 1, a);
            if (a[l] > a[r]) {
                swap(l, r, a);
            }
            if (a[l + 1] > a[r]) {
                swap(l + 1, r, a);
            }
            if (a[l] > a[l + 1]) {
                swap(l, l + 1, a);
            }

            // performing division
            // barrier is a[l + 1], i.e. median among a[l], a[l + 1], a[r]
            int i = l + 1,
                    j = r;
            final int cur = a[l + 1];
            for (;;) {
                while (a[++i] < cur) ;
                while (a[--j] > cur) ;
                if (i > j) {
                    break;
                }
                swap(j, i, a);
            }

            // inserting the barrier
            a[l + 1] = a[j];
            a[j] = cur;

            // we continue to work in that part, which must contain the required element
            if (j >= k) {
                r = j - 1;
            }
            if (j <= k) {
                l = i;
            }
        }
    }*/
}

//class Pair {

/*SORTING PAIRS BY COMPARING Y OF PAIR.
    Pair[] p=new Pair[n];
    p[i]=new Pair(i,ni());
    Arrays.sort(p,( p1, p2) -> {return p1.y-p2.y;});
 */
 /*int x;
    int y;

    Pair(int u, int v) {
        x = u;
        y = v;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) {
            return true;
        }
        if (!(o instanceof Pair)) {
            return false;
        }
        Pair key = (Pair) o;
        return x == key.x && y == key.y;
    }

    @Override
    public int hashCode() {
        int result = x;
        result = 31 * result + y;
        return result;
    }

}*/