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
import sys
stdin = sys.stdin

sys.setrecursionlimit(10**5)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

n = ni()
s = " " + ns()
q = ni()
k = list(li())

def find_ans(k:int):
    cnt = [0,0,0]
    ans = 0
    
    for i in range(1, n+1):
        
        if s[i] == 'C':
            ans += cnt[2]

        if i-k+1 > 0:
            if s[i-k+1] == 'D':
                cnt[0] -= 1
                cnt[2] -= cnt[1]
                
            elif s[i-k+1] == 'M':
                cnt[1] -= 1
                
        if s[i] == 'D':
            cnt[0] += 1    
        
        elif s[i] == 'M':
            cnt[2] += cnt[0]
            cnt[1] += 1

            
    return ans
        
for ki in k:
    print(find_ans(ki))