In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.
The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.

Given a string S of length N and an integer k (k \geq 3), he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:

* 0 \leq a < b < c \leq N - 1
* S[a] = `D`
* S[b] = `M`
* S[c] = `C`
* c-a < k



Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.

For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).

Constraints

* 3 \leq N \leq 10^6
* S consists of uppercase English letters
* 1 \leq Q \leq 75
* 3 \leq k_i \leq N
* All numbers given in input are integers

Input

Input is given from Standard Input in the following format:


N
S
Q
k_{0} k_{1} ... k_{Q-1}


Output

Print Q lines. The i-th line should contain the k_i-DMC number of the string S.

Examples

Input

18
DWANGOMEDIACLUSTER
1
18


Output

1


Input

18
DDDDDDMMMMMCCCCCCC
1
18


Output

210


Input

54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40


Output

0
1
2


Input

30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20


Output

10
52
110
140
n = int(input()) + 1
s = '?' + input()
q = int(input())
k = list(map(int,input().split()))

p_d = []
cnt_m = [0]*n
cnt_c = [0]*n

for i,si in enumerate(s[1:],1):
    cnt_m[i] = cnt_m[i-1]
    cnt_c[i] = cnt_c[i-1]
    if(si=='D'):
        p_d.append(i)
    elif(si=='M'):
        cnt_m[i] += 1
    elif(si=='C'):
        cnt_c[i] += 1

left_c = [0]*n
for i in range(1,n):
    left_c[i] = left_c[i-1]
    if(s[i]=='M'):
        left_c[i] += cnt_c[i]

# print(cnt_m)
# print(cnt_c)
# print(left_c)

ans = []
for ki in k:
    tmp = 0
    for di in p_d:
        x = cnt_m[min(di+ki-1,n-1)] - cnt_m[di-1]
        y = cnt_c[min(di+ki-1,n-1)]
        tmp += x*y - left_c[min(di+ki-1,n-1)] + left_c[di-1]
    ans.append(tmp)

print('\n'.join(map(str,ans)))