Problem

Select 3 grid points from the closed rectangular section surrounded by points (0,0), (n-1,0), (0, m-1), (n-1, m-1) and make a triangle. make.
At this time, the absolute difference between the number of triangles containing the point (x1, y1) but not the point (x2, y2) and the number of triangles containing the point (x2, y2) but not the point (x1, y1) Find the value.
Those whose points are on the border are considered to be included in the triangle.
However, if the three points are aligned in a straight line, it is not considered to be a triangle.

Constraints

* 2 ≤ n, m ≤ 500
* 0 ≤ x1, x2 ≤ n-1
* 0 ≤ y1, y2 ≤ m-1
* (x1, y1) ≠ (x2, y2)

Input


n m x1 y1 x2 y2


All inputs are given as integers.
N, m, x1, y1, x2, y2 are given on one line, separated by blanks.

Output

Output the absolute value of the difference on one line.

Examples

Input

2 2 0 0 0 1


Output

0


Input

3 2 1 0 0 0


Output

3
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double EPS=1e-10,pi=M_PI;
typedef complex<double> P;
double cross(const P& a, const P& b) {return imag(conj(a)*b);}
double dot(const P& a, const P& b) {return real(conj(a)*b);}
bool cmp(P a, P b) {
  double at=atan2(a.imag(),a.real()),at2=atan2(b.imag(),b.real());
  return at!=at2?at<at2:max(abs(a.real()),abs(a.imag()))<max(abs(b.real()),abs(b.imag()));
}
int ccw(P a, P b, P c) {
  b-=a;c-=a;
  if(cross(b,c)>EPS)return +1;
  if(cross(b,c)<-EPS)return -1;
  if(dot(b,c)<-EPS)return +2;
  if(norm(b)<norm(c))return -2;
  return 0;
}

int main() {
  ll n,m,x,y,sum[2];
  cin >> n >> m;
  for(int k=0; k<2; k++) {
    cin >> x >> y;
    if(x==n-1 && y==m-1) x=0,y=0;
    vector<P> v;
    map<double,ll> d;
    for(ll i=0; i<n; i++) {
      for(ll j=0; j<m; j++) {
        if(i-x || j-y) v.push_back(P(i-x,j-y));
      }
    }
    sort(v.begin(),v.end(),cmp);
    sum[k]=n*m*(n*m-1)*(n*m-2)/6;
    ll cnt=0,j=v.size()-1;
    while(j && (ccw(v[0],P(0,0),v[j])==1 || ccw(v[0],P(0,0),v[j])==0)) {
      cnt++;
      j--;
    }
    j=(j+1)%v.size();
    for(int i=0; i<v.size(); i++) {
      while(i!=j && ccw(v[i],P(0,0),v[j])!=1 && ccw(v[i],P(0,0),v[j])!=0) {
        cnt--;
        j=(j+1)%v.size();
      }
      sum[k]-=cnt*(cnt-1)/2;
      double at=atan2(v[i].imag(),v[i].real()),xx=v[i].real(),yy=v[i].imag();
      if(xx) xx=-xx;
      if(yy) yy=-yy;
      double at2=atan2(yy,xx);
      sum[k]-=d[at2]*(d[at2]+1)/2+d[at]*(d[at2]+1);
      cnt++;
      d[at]++;
    }
  }
  cout << abs(sum[0]-sum[1]) << endl;
  return 0;
}
