CodeForces

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 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|.

Example

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.

用第一串a对第二串b（b中长度为len（a）的子串）不断进行比较，数字相同答案不变，不同则答案加一。

若从一般的方式直接比较，显然会超时，但注意只有0和1两种状态，所以，只要记录a串的0的数量和1的数量，是可以避免对b串的重复操作。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
using namespace std;
const int MAX=2e5+10;
const double MIN=1e-6;
#define INF 0x7fffffff
#define ll long long
#define FOR(i,n) for(i=1;i<=n;i++)
#define mst(a) memset(a,0,sizeof(a))
#define mstn(a,n) memset(a,n,sizeof(a))
//struct num{int a,b;}a[1005];
//bool cmp(const num &x, const num &y){return x.a>y.a;}

int main()
{
    string a,b;
    int i;
    cin>>a;
    cin>>b;
    ll ans=0;
    int d=b.size()-a.size()+1;
    ll s=0;
    for(i=0;i<d;i++)
    {
        s+=b[i]-'0';
    }
    for(i=0;i<a.size();i++)
    {
        if(a[i]=='0') ans+=s;
            else ans+=d-s;
        s+=(b[d+i]-'0')-(b[i]-'0');
    }
    cout<<ans<<endl;
    return 0;
}