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.
2
2 2
yz
1 1
2 2
yz
1 1
 1
11 2
thisisatest
3 5

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 1 

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 1 

 1 0 0 0 1 0 0 3 5 0 0 0 0 0 0 0 0 0 4 5 0 0 0 0 0 0 




# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
# sys.setrecursionlimit(10**5)

	
def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())
def bi(x):
	return bin(x)[2:]










from copy import deepcopy
def solution(s,n,p,m):
	s = [0] + s
	dp = [[0 for _ in range(26)] for _ in range(n+1)]
	p.sort()
	for i in range(1,n+1):
		for j in range(26):
			dp[i][j] = dp[i-1][j] + int(j == (ord(s[i])-ord('a')))
	#total m+1 tries
	res = [x for x in dp[n]]
	for i in range(m):
		x=p[i]
		for j in range(26):
			res[j]+=dp[x][j]
	return ' '.join(map(str,res))





def main():
	T = read_int()
	for i in range(T):
		n,m = read_int_array()
		s = read_string()
		p = read_int_array()
		x = solution(s,n,p,m)

		if 'xrange' not in dir(__builtins__):
			print(x)
		else:
			print >>output,str(x)# "Case #"+str(i+1)+':',
	if 'xrange' in dir(__builtins__):
		print(output.getvalue())
		output.close()

if 'xrange' in dir(__builtins__):
	import cStringIO
	output = cStringIO.StringIO()
#example usage:
#    for l in res:
#       print >>output, str(len(l)) + ' ' +  ' '.join(l)

if __name__ == '__main__':
	main()



'''stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* BITS - THINK HOW TO MASK PROPERLY
	* PERMUTATIONS - PARITY AND CYCLES
	* Think simple, if it becomes over complicated, try to look at it from a different perspective.
	* Have fun!!!
	* TRY FIXING SOMETHING, and then maybe binary search around it.
	* Remember heaps. 
	* Remember how to add a value to a segment when using prefix sum.
		suppose you have an array[1,2,3,4,5] and you want to add 3 to array[1:4]. Then just 
		add 3 to A[1], and decrease 3 from A[4]. Let's look at what happens:
		original prefixsums is 					 [1,3,6,10,15]
		array -> [1,5,3,4,2] and prefix sums are [1,6,9,13,15] 
		As you see, exactly +3 in A[1:4]
		*** The previous method can help checking how many x,y you can choose to get s=x+y from two arrays.
'''
'''
binary search 
	while(r - l > 1) {
		ll mid = l + (r - l) / 2;
		solve(mid);
		ll sum = 0;
		for (int i = 0; i < n; i++)
			sum += b[i];
		if (sum <= k)
			r = mid;
		else
			l = mid;
	}
'''
