1186_C. 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| ≤ |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.
```python
a=input()
b=input()
A=[]
al=len(a)
bl=len(b)
A.append(0)
count=0
for x in a:
    if x=='1':
        count+=1
    A.append(count)
evencount=0
bcount=0
for x in b:
    if x=='1':
        bcount+=1
for i in range((al+1)-bl):
    acount=A[i+bl]-A[i]
    if (acount%2)==(bcount%2):
        evencount+=1
print(evencount)

```