Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

n, P = map(int, input().split())
S = input()
ans = 0
#----------------------------------------------
# 1.
# int(S[a:b]) * 10 ** (n - b) == (int(S[a: n]) - int(S[b: n])) で表される
# 例
# S = 1234
# a = 1, b = 3 のとき
# 23 = (234 - 4) // 10 ** (4 - 3)
#    =  230 // 10
#    =  23
#
# 2.
# Pは素数のため　P = 2, 5　のとき以外は　１０の約数でない
# A * (10 ** x) が Pで割れるとき　A は P の倍数　(x は任意定数)
#
# 3.
# 1. の式を mod P で考える
# int(S[a: b]) * 10 ** (n - b) == (int(S[a: n]) - int(S[b: n]))
# ⇒　(S[a: b] * 10 ** (n - b)) % P == (S[a: n] - S[b: n]) % P  (mod P)
# ⇒　(S[a: b] % P) * (10 ** (n - b) % P) == (S[a: n] % P) - (S[b: n] % P)  (mod P)
# 2.より P = 2, 5以外のときは 10 ** (n - b) % P != 0となるため
# (S[a: n] % P) - (S[b: n] % P)　= 0　すなわち　(S[a: n] % P) == (S[b: n] % P)のとき
# S[a:b] % P　= 0である
#
# これより
# S[i: n] % P (0 <= i < n) を求めれば
# 以下のコードのようにS[i: j] % P を求めることができる (0 <= i < n, i <= j < n)
#----------------------------------------------
#まずS[i: n] % P (0 <= i < n) を求める (remain に格納する)
remain = [0] * P
remain[0] += 1
d = 1
left = 0
for i in range(n):
    print(left)
    left = (left + int(S[n - i - 1]) * d) % P
    remain[left] += 1
    d = d * 10 % P
# print(remain)
for r in remain:
    ans += r * (r - 1) // 2
#----------------------------------------------
# 4.P = 2, 5　のときについて考える
#     1. P = 2 のとき
#         ある数の下１桁が2の倍数のとき、その数は2の倍数であるため
#         S[i] % 2 == 0 のとき
#         S[h:i] % 2 == 0 (0 <= h <= i)
#         よって以下のようなコードで答えを求めることができる
#----------------------------------------------
if P == 2:
    ans = 0
    for i in range(n):
        if int(S[i]) % 2 == 0:
            ans += i + 1
    print(ans)
#----------------------------------------------
    # 2. P = 5 のときもP = 2 のときと同様に考えれば良いため
#----------------------------------------------
if P == 5:
    ans = 0
    for i in range(n):
      if int(S[i]) % 5 == 0:
          ans += i + 1
    print(ans)
if P != 2 and P != 5:
    print(ans)