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.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static ArrayList<Integer> adj[];
	static PrintWriter out = new PrintWriter(System.out);
	public static int mod = 1000000007;

	static int notmemo[][][][];
	static int k;
	static int a[];
	static int b[];
	static int m;
	static char c[];

	static int trace[];
	static int h;
	static int x;
	static int ans1;
	static int ans2;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int t=sc.nextInt();
		label:while(t-->0) {
			int n=sc.nextInt();
			long m=sc.nextInt();
			String s=sc.nextLine();
			char c[]=s.toCharArray();
		    long count[]=new long[26];
		 long p[]=new long[(int) (m+1)];
			for (int i = 0; i < p.length-1; i++) {
				p[i]=sc.nextInt()-1;
			}
			p[(int) m]=n;
			schuffle(p);
			Arrays.sort(p);
			long start=p[0];
			int j=0;
			for (int i = 0; i < c.length; i++) {
				if(i<=start) {
				count[c[i]-'a']+=((p.length-j));	
				//System.out.println(c[i]);
				}
				while(i==start) {
					j++;
					if(j>=m+1)
						break;
					start=p[j];
					
				}
				}
for (int i = 0; i < count.length; i++) {
	System.out.print(count[i]+" ");
}
System.out.println();
		}

		

	}
	static class book implements Comparable<book> {
		int idx;
		int score;

		public book(int i, int s) {
			idx = i;
			score = s;
		}

		public int compareTo(book o) {
			return o.score - score;
		}
	}

	static class library implements Comparable<library> {
		int numofbooks;
		int signup;
		int shiprate;
		int idx;

		public library(int a, int b, int c, int idx) {
			numofbooks = a;
			signup = b;
			shiprate = c;
			this.idx = idx;
		}

		@Override
		public int compareTo(library o) {
		if(signup==o.signup) {
			  return o.numofbooks-numofbooks;
		}
		return signup - o.signup;
		
		}
	}

	static long dp(int i, int state) {
		if (i < 0) {
			return state;
		}
		if (memo[i][state] != -1)
			return memo[i][state];

		int x = Integer.parseInt(c[i] + "");
		x += state;
		int newc = x / 10;
		x = x % 10;

		return memo[i][state] = Math.min(dp(i - 1, newc) + x, x == 0 ? (long) 1e15 : dp(i - 1, newc + 1) + 10 - x);
	}

	static boolean isPal(String s) {
		int j = s.length() - 1;
		c = s.toCharArray();
		Arrays.sort(c);
		for (int i = 0; i < c.length; i++) {

		}
		return true;
	}

	static boolean dfs(int u, int color) {
		vis[u] = true;

		boolean f = true;
		for (int v : adj[u]) {
			if (u != ans1 && a[v] != a[u] && v != ans1) {
				// System.out.println(u+" "+v+" "+a[u]+" "+a[v]+" "+ans1);
				return false;
			}
			if (!vis[v]) {
				f &= dfs(v, a[v]);

			}
		}
		return f;

	}

	static String s;

	static boolean isOn(int S, int j) {
		return (S & 1 << j) != 0;
	}

	static int y, d;

	static void extendedEuclid(int a, int b) {
		if (b == 0) {
			x = 1;
			y = 0;
			d = a;
			return;
		}
		extendedEuclid(b, a % b);
		int x1 = y;
		int y1 = x - a / b * y;
		x = x1;
		y = y1;
	}

	static boolean f = true;

	static class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;

		SegmentTree(int[] in) {
			array = in;
			N = in.length - 1;
			sTree = new int[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N << 1];
			build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;
			sTree[index] += val;
			while (index > 1) {
				index >>= 1;
				sTree[index] = Math.max(sTree[index << 1], sTree[index << 1 | 1]);
			}
		}

		void update_range(int i, int j, int val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, int val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] += (e - b + 1) * val;
				lazy[node] += val;
			} else {
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		int query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		int query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			// propagate(node, b, mid, e);
			int q1 = query(node << 1, b, mid, i, j);
			int q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return Math.max(q1, q2);

		}
	}

	static long memo[][];

	static long nCr1(int n, int k) {
		if (n < k)
			return 0;
		if (k == 0 || k == n)
			return 1;
		if (comb[n][k] != -1)
			return comb[n][k];
		if (n - k < k)
			return comb[n][k] = nCr1(n, n - k);
		return comb[n][k] = ((nCr1(n - 1, k - 1) % mod) + (nCr1(n - 1, k) % mod)) % mod;
	}

	static class UnionFind {
		int[] p, rank, setSize;
		int numSets;
		int max[];

		public UnionFind(int N) {
			max = new int[N];
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {
				p[i] = i;
				setSize[i] = 1;
				max[i] = i;
			}
		}

		public int findSet(int i) {
			return p[i] == i ? i : (p[i] = findSet(p[i]));
		}

		public boolean isSameSet(int i, int j) {
			return findSet(i) == findSet(j);
		}

		public void unionSet(int i, int j) {
			if (isSameSet(i, j))
				return;
			numSets--;
			int x = findSet(i), y = findSet(j);
			if (rank[x] > rank[y]) {
				p[y] = x;
				setSize[x] += setSize[y];
				max[x] = Math.max(max[x], max[y]);

			} else {
				p[x] = y;
				setSize[y] += setSize[x];
				if (rank[x] == rank[y])
					rank[y]++;
				max[y] = Math.max(max[x], max[y]);

			}
		}

		private int getmax(int j) {
			return max[findSet(j)];
		}

		public int numDisjointSets() {
			return numSets;
		}

		public int sizeOfSet(int i) {
			return setSize[findSet(i)];
		}
	}

	/**
	 * private static void trace(int i, int time) { if(i==n) return;
	 * 
	 * 
	 * long ans=dp(i,time);
	 * if(time+a[i].t<a[i].burn&&(ans==dp(i+1,time+a[i].t)+a[i].cost)) {
	 * 
	 * trace(i+1, time+a[i].t);
	 * 
	 * l1.add(a[i].idx); return; } trace(i+1,time);
	 * 
	 * }
	 **/

	static class incpair implements Comparable<incpair> {
		int a;
		long b;
		int idx;

		incpair(int a, long dirg, int i) {
			this.a = a;
			b = dirg;
			idx = i;
		}

		public int compareTo(incpair e) {
			return (int) (b - e.b);
		}
	}

	static class decpair implements Comparable<decpair> {
		int a;
		long b;
		int idx;

		decpair(int a, long dirg, int i) {
			this.a = a;
			b = dirg;
			idx = i;
		}

		public int compareTo(decpair e) {
			return (int) (e.b - b);
		}
	}

	static long allpowers[];

	static class Quad implements Comparable<Quad> {
		int u;
		int v;
		char state;
		int turns;

		public Quad(int i, int j, char c, int k) {
			u = i;
			v = j;
			state = c;
			turns = k;
		}

		public int compareTo(Quad e) {
			return (int) (turns - e.turns);
		}

	}

	static long dirg[][];
	static Edge[] driver;

	static int n;

	static class Edge implements Comparable<Edge> {
		int node;
		long cost;

		Edge(int a, long dirg) {
			node = a;
			cost = dirg;
		}

		public int compareTo(Edge e) {
			return (int) (cost - e.cost);
		}
	}

	static long manhatandistance(long x, long x2, long y, long y2) {
		return Math.abs(x - x2) + Math.abs(y - y2);
	}

	static long fib[];

	static long fib(int n) {
		if (n == 1 || n == 0) {
			return 1;
		}
		if (fib[n] != -1) {
			return fib[n];
		} else
			return fib[n] = ((fib(n - 2) % mod + fib(n - 1) % mod) % mod);
	}

	static class Pair {
		int a, b;

		Pair(int x, int y) {
			a = x;
			b = y;
		}

	}

	static long[][] comb;

	static class Triple implements Comparable<Triple> {

		int l;
		int r;
		long cost;
		int idx;

		public Triple(int a, int b, long l1, int l2) {
			l = a;
			r = b;
			cost = l1;
			idx = l2;
		}

		public int compareTo(Triple x) {
			if (l != x.l || idx == x.idx)
				return l - x.l;
			return -idx;
		}

	}

	static TreeSet<Long> primeFactors(long N) // O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			while (N % p == 0) {
				factors.add((long) p);
				N /= p;
			}
			if (primes.size() > idx + 1)
				p = primes.get(++idx);
			else
				break;
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];

	/**
	 * static int bfs(int s) { Queue<Integer> q = new LinkedList<Integer>();
	 * q.add(s); int count=0; int maxcost=0; int dist[]=new int[n]; dist[s]=0;
	 * while(!q.isEmpty()) {
	 * 
	 * int u = q.remove(); if(dist[u]==k) { break; } for(Pair v: adj[u]) {
	 * maxcost=Math.max(maxcost, v.cost);
	 * 
	 * 
	 * 
	 * if(!visited[v.v]) {
	 * 
	 * visited[v.v]=true; q.add(v.v); dist[v.v]=dist[u]+1; maxcost=Math.max(maxcost,
	 * v.cost); } }
	 * 
	 * } return maxcost; }
	 **/
	public static boolean FindAllElements(int n, int k) {
		int sum = k;
		int[] A = new int[k];
		Arrays.fill(A, 0, k, 1);

		for (int i = k - 1; i >= 0; --i) {

			while (sum + A[i] <= n) {

				sum += A[i];
				A[i] *= 2;
			}
		}
		if (sum == n) {
			return true;
		} else
			return false;
	}

	static boolean vis2[][];

	static boolean f2 = false;

	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r) // C(p x r) = A(p x q) x (q x r) -- O(p x q x
																			// r)
	{
		long[][] C = new long[p][r];
		for (int i = 0; i < p; ++i) {
			for (int j = 0; j < r; ++j) {
				for (int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j] + (a2[i][k] % mod * b[k][j] % mod)) % mod;
					C[i][j] %= mod;
				}

			}
		}
		return C;
	}

	static ArrayList<Pair> a1[];
	static int memo1[];

	static boolean vis[];
	static TreeSet<Integer> set = new TreeSet<Integer>();

	static long modPow(long ways, long count, long mod) // O(log e)
	{
		ways %= mod;
		long res = 1;
		while (count > 0) {
			if ((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res % mod;
	}

	static long gcd(long ans, long b) {
		if (b == 0) {
			return ans;
		}
		return gcd(b, ans % b);
	}

	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;

	static void sieve(int N) // O(N log log N)
	{
		isComposite = new int[N + 1];
		isComposite[0] = isComposite[1] = 1; // 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l = new ArrayList<>();
		valid = new int[N + 1];
		for (int i = 2; i <= N; ++i) // can loop till i*i <= N if primes array is not needed O(N log log sqrt(N))
			if (isComposite[i] == 0) // can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i] = 1;

				for (int j = i * 2; j <= N; j += i) { // j = i * 2 will not affect performance too much, may alter in
														// modified sieve
					isComposite[j] = 1;

				}
			}

		for (int i = 0; i < primes.size(); i++) {
			for (int j : primes) {
				if (primes.get(i) * j > N) {
					break;
				}
				valid[primes.get(i) * j] = 1;
			}
		}
	}

	public static long[] schuffle(long[] p) {
		for (int i = 0; i < p.length; i++) {
			long x = (long) (Math.random() * p.length);
			long temp = p[(int) x];
			p[(int) x] = p[i];
			p[i] = temp;
		}
		return p;
	}

	static int V;
	static long INF = (long) 1E16;

	static class Edge2 {
		int node;
		long cost;
		long next;

		Edge2(int a, int c, Long long1) {
			node = a;
			cost = long1;
			next = c;
		}

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

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

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

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

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

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

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}