Codeforces Round #416 (Div. 2) has been moved to start on 27.05.2017 09:35 (UTC).
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B. Hamming Distance Sum

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputGenos needs your help. He was asked to solve the following programming problem by Saitama:

The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s*_{i} is the *i*-th character of *s* and *t*_{i} is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0 - 0| + |0 - 1| + |1 - 1| + |1 - 0| = 0 + 1 + 0 + 1 = 2.

Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.

Input

The first line of the input contains binary string *a* (1 ≤ |*a*| ≤ 200 000).

The second line of the input contains binary string *b* (|*a*| ≤ |*b*| ≤ 200 000).

Both strings are guaranteed to consist of characters '0' and '1' only.

Output

Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.

Examples

Input

01

00111

Output

3

Input

0011

0110

Output

2

Note

For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.

The second sample case is described in the statement.

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

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