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| ≤ |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.

Input

The first line contains a binary string a (1 ≤ |a| ≤ 10^6) — the first string.

The second line contains a binary string b (1 ≤ |b| ≤ |a|) — the second string.

Output

Print one number — the answer.

Examples

Input


01100010
00110


Output


3


Input


1010111110
0110


Output


4

Note

The first example is explained in the legend.

In the second example, there are five substrings that satisfy us: 1010, 0101, 1111, 1111.
#########################################################################################################\
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###################################The_Apurv_Rathore#####################################################
#########################################################################################################
#########################################################################################################
from decimal import *
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math 

sys.setrecursionlimit(100000)

alphabets = list('abcdefghijklmnopqrstuvwxyz')


def isPrime(x):
    for i in range(2,x):
        if i*i>x:
            break
        if (x%i==0):
            return False
    return True
def ncr(n, r, p):  
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,p - 2, p)) % p
def primeFactors(n): 
    l = []
    while n % 2 == 0: 
        l.append(2)
        n = n / 2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            l.append(int(i))
            n = n / i 
    if n > 2: 
        l.append(n)
    c = dict(Counter(l))
    return list(set(l))
    # return c

def power(x, y, p) : 
	res = 1
	x = x % p 
	if (x == 0) : 
		return 0
	while (y > 0) : 
		if ((y & 1) == 1) : 
			res = (res * x) % p 
		y = y >> 1	 # y = y/2 
		x = (x * x) % p 		
	return res 

#____________________GetPrimeFactors in log(n)________________________________________
def sieveForSmallestPrimeFactor():
    MAXN = 100001
    spf = [0 for i in range(MAXN)]
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i
    for i in range(4, MAXN, 2):
        spf[i] = 2
    for i in range(3, math.ceil(math.sqrt(MAXN))):
        if (spf[i] == i):
            for j in range(i * i, MAXN, i): 
                if (spf[j] == j):
                    spf[j] = i
    return spf
def getPrimeFactorizationLOGN(x):
    spf = sieveForSmallestPrimeFactor()
    ret = list()
    while (x != 1):
        ret.append(spf[x])
        x = x // spf[x]  
    return ret
#____________________________________________________________



def SieveOfEratosthenes(n): 
    #time complexity = nlog(log(n))
    prime = [True for i in range(n+1)]
    p = 2
    pr = []
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    for i in range(2,len(prime)):
        if prime[i]:
            pr.append(i)
    return pr
def si():
    return input()
def divideCeil(n,x):
    #return (n+1)//x
    if (n%x==0):
        return n//x
    return n//x+1
def ii():
    return int(input())
def li():
    return list(map(int,input().split()))

#__________________________TEMPLATE__________________OVER_______________________________________________________


if(os.path.exists('input.txt')):
    sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
# else:
#     input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline


def solve():

    a = list(si())
    b = list(si())
    cntb = b.count('1')
    ans = 0
    i = 0
    s = a[:len(b)].count('1')
    for i in range(len(a)):
        # print(s)
        if s%2==cntb%2:
            ans+=1
        if i+len(b)>=len(a):
            break
        if a[i+len(b)]=='1':
            s+=1
        if a[i]=='1':
            s-=1
    print(ans)

t = 1
# t = ii()
for _ in range(t):
    solve()
