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 PYTHON3):

Solution:
#_______________________________________________________________#
def fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): #return 0 if all are greater or equal to
	answer = -1 
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): #return n-1 if all are small or equal
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub):
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans

def sieve_of_eratosthenes(n):
	ans = []
	arr = [1]*(n+1)
	arr[0],arr[1], i = 0, 0, 2
	while(i*i <= n):
		if arr[i] == 1:
			j = i+i
			while(j <= n):
				arr[j] = 0
				j += i
		i += 1
	for k in range(n):
		if arr[k] == 1:
			ans.append(k)
	return ans
def nc2(x):
	if x == 1:
		return 0
	else:
		return x*(x-1)//2
def kadane(x):
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = max(sum_so_far,current_sum)
	return sum_so_far
def mex(li):
	check = [0]*1001
	for i in li:
		check[i] += 1
	for i in range(len(check)):
		if check[i] == 0:
		 return i 
def sumdigits(n):
	ans = 0
	while(n!=0):
		ans += n%10
		n //= 10
	return ans

#_______________________________________________________________#
'''
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'''
from math import *
from collections import deque
max_steps = 100005
for _ in range(int(input())):
#for _ in range(1):
	#n = int(input())
	n,m = map(int,input().split())
	st = input()
	a = list(map(int,input().split()))
	#b = list(map(int,input().split()))
	li = [0]*(len(st)+1)
	li[0] = m+1
	for i in a:
		li[i] -= 1
	for i in range(1,len(li)):
		li[i] += li[i-1]
	char = [0]*26
	for i in range(len(st)):
		char[ord(st[i])-97] += li[i]
	print(*char)