Genos 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 si is the i-th character of s and ti 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|.

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.

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

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 is1 + 0 + 1 + 1 = 3.

The second sample case is described in the statement.

题意：输入两个只包含1和0的字符串a,b（1≤|a| ≤ |b| ≤ 200 000）。a与b中所有相等长度的连续子串进行比较，结果为每位数差的绝对值之和。输出所有结果的和。

样例1：a是01，b是00111；01与00（b中的第1位和第2位）比较，01与01（b中的第2位和第3位）比较，01与11（b中的第3位和第4位）比较，01与11（b中的第4位和第5位）比较。每一个比较的结果分别为1,0,1,1。输出结果的和3。

思路分析：a中的第1个数要和b中的第1,2,3,4个数进行比较，a中的第2个数要和b中的第2,3,4,5个数进行比较。a中的每一个数都比较了4次，设字符串a的长度为lena，b的长度为lenb，那就是要比较compare=lenb-lena+1次。每一次比较只要统计当前区间 [ b[i]，b[i+compare]）b为1的个数sum，如果a[i]=1,就加上compare-sum;如果a[i]=0，就加上sum。第一次sum的值就是for(i=0;i<compare;i++) if(b[i]==1) sum++;以后每次比较完之后就只要if(b[i]==1) sum--;if(b[i+compare]==1) sum++;

#include<iostream>

#include<cstdio>

using namespace std;

int main()

{

    string a,b;

    int na,nb,compare;

    cin>>a>>b;

    na=a.size();

    nb=b.size();

    compare=nb-na+;

    int i;

    __int64 sum=,ans=;

    for(i=; i<compare; i++)

        if(b[i]=='')sum++;

    for(i=; i<na; i++)

    {

        if(a[i]=='') ans+=compare-sum;

        else ans+=sum;

        if(b[i]=='')sum--;

        if(b[i+compare]=='')sum++;

    }

    cout<<ans<<endl;

}