Vus the Cossack and Strings
Vus the Cossack has two binary strings, that is, strings that consist only of "0" and "1". We call these strings $$$a$$$ and $$$b$$$. It is known that $$$|b| \leq |a|$$$, that is, the length of $$$b$$$ is at most the length of $$$a$$$.The Cossack considers every substring of length $$$|b|$$$ in string $$$a$$$. Let's call this substring $$$c$$$. He matches the corresponding characters in $$$b$$$ and $$$c$$$, after which he counts the number of positions where the two strings are different. We call this function $$$f(b, c)$$$.For example, let $$$b = 00110$$$, and $$$c = 11000$$$. In these strings, the first, second, third and fourth positions are different.Vus the Cossack counts the number of such substrings $$$c$$$ such that $$$f(b, c)$$$ is even.For example, let $$$a = 01100010$$$ and $$$b = 00110$$$. $$$a$$$ has four substrings of the length $$$|b|$$$: $$$01100$$$, $$$11000$$$, $$$10001$$$, $$$00010$$$.  $$$f(00110, 01100) = 2$$$; $$$f(00110, 11000) = 4$$$; $$$f(00110, 10001) = 4$$$; $$$f(00110, 00010) = 1$$$. Since in three substrings, $$$f(b, c)$$$ is even, the answer is $$$3$$$.Vus can not find the answer for big strings. That is why he is asking you to help him.
The first example is explained in the legend.In the second example, there are five substrings that satisfy us: $$$1010$$$, $$$0101$$$, $$$1111$$$, $$$1111$$$.
The first line contains a binary string $$$a$$$ ($$$1 \leq |a| \leq 10^6$$$) — the first string.The second line contains a binary string $$$b$$$ ($$$1 \leq |b| \leq |a|$$$) — the second string.
Print one number — the answer.
def solve(a, b):
    len_a = len(a)
    len_b = len(b)
    bits_xored_subarr_a = 0
    bits_xored_b = 0
    for i in range(0, len_b):
        if a[i] == '1':
            bits_xored_subarr_a ^= 1
        if b[i] == '1':
            bits_xored_b ^= 1
    count = 0
    for i in range(len_b, len_a):
        if bits_xored_subarr_a == bits_xored_b:
            count += 1
        if a[i] == '1':
            bits_xored_subarr_a ^= 1
        if a[i - len_b] == '1':
            bits_xored_subarr_a ^= 1
    if bits_xored_subarr_a == bits_xored_b:
        count += 1
    return count
if __name__ == '__main__':
    a = input()
    b = input()
    print(solve(a, b))
""""""

def solve(a, b):
    len_a = len(a)
    len_b = len(b)
    bits_xored_subarr_a = 0
    bits_xored_b = 0
    for i in range(0, len_b):
        if a[i] == '1':
            bits_xored_subarr_a ^= 1
        if b[i] == '1':
            bits_xored_b ^= 1
    count = 0
    for i in range(len_b, len_a):
        if bits_xored_subarr_a == bits_xored_b:
            count += 1
        if a[i] == '1':
            bits_xored_subarr_a ^= 1
        if a[i - len_b] == '1':
            bits_xored_subarr_a ^= 1
    if bits_xored_subarr_a == bits_xored_b:
        count += 1
    return count
if __name__ == '__main__':
    a = input()
    b = input()
    print(solve(a, b))
import math

def main():
    a = input()
    b = input()
    c = a[:len(b)]
    cnt = 0
    cntd = 0
    ans = 0
    for i in c:
        if i == '1':
            cnt += 1
    for i in b:
        if i == '1':
            cntd += 1
    if cnt % 2 == cntd % 2:
        ans += 1
    for i in range(len(b), len(a)):
        if a[i] == '1':
            cnt += 1
        if a[i - len(b)] == '1':
            cnt -= 1
        if cnt % 2 == cntd % 2:
            ans += 1
    print(ans)
main()