Two coordinates (a1, a2) and (b1, b2) on a two-dimensional grid of r × c are given. The cost of moving from a cell (e, f) to one of the cells (e + 1, f), (e-1, f), (e, f + 1), (e, f-1) is 1. And. You can also move between (e, c-1) and (e, 0), and between (r-1, f) and (0, f) at a cost of 1. At this time, find the number of routes that can be moved from the first coordinate to the second coordinate at the shortest cost.



Input

The input is given in the following format.


r c a1 a2 b1 b2


Input meets the following constraints
1 ≤ r, c ≤ 1,000
0 ≤ a1, b1 <r
0 ≤ a2, b2 <c

Output

Output the remainder of the answer value divided by 100,000,007.

Examples

Input

4 4 0 0 3 3


Output

2


Input

4 4 0 0 1 1


Output

2


Input

2 3 0 0 1 2


Output

4


Input

500 500 0 0 200 200


Output

34807775
#include <iostream>
#include <cstdio>
#include <vector>
#include <list>
#include <cmath>
#include <fstream>
#include <algorithm>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <complex>
#include <iterator>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <stack>
#include <climits>
#include <deque>
#include <bitset>
#include <cassert>
#include <ctime>
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
const int dy[]={-1,0,1,0},dx[]={0,1,0,-1};
// adjust problem by problem
const double EPS=1e-8;
const double PI=acos(-1.0);
#ifdef __GNUC__
int popcount(int n){return __builtin_popcount(n);}
int popcount(ll n){return __builtin_popcountll(n);}
#endif
#ifndef __GNUC__
template<class T> int popcount(T n){int cnt=0;while(n){if(n%2)cnt++;n/=2;}return cnt;}
#endif
template<class T>int SIZE(T a){return a.size();}
template<class T>string IntToString(T num){string res;stringstream ss;ss<<num;return ss.str();}
template<class T>T StringToInt(string str){T res=0;for(int i=0;i<SIZE(str);i++)res=(res*10+str[i]-'0');return res;}
template<class T>T gcd(T a,T b){if(b==0)return a;return gcd(b,a%b);}
template<class T>T lcm(T a,T b){return a/gcd(a,b)*b;}
bool EQ(double a,double b){return abs(a-b)<EPS;}
void fastStream(){cin.tie(0);std::ios_base::sync_with_stdio(0);}
vector<string> split(string str,char del){
  vector<string> res;
  for(int i=0,s=0;i<SIZE(str);i++){
    if(str[i]==del){if(i-s!=0)res.push_back(str.substr(s,i-s));s=i+1;}
    else if(i==SIZE(str)-1){res.push_back(str.substr(s));}
  }
  return res;
}

int H,W;
int sx,sy,gx,gy;

const int mod=100000007;
const int MAX_FACT=10000;

ll fact[MAX_FACT];

ll pow_mod(ll a,ll n){
  ll cur=1;
  ll mul=a;
  while(n){
    if(n%2)cur=(cur*mul)%mod;
    mul=(mul*mul)%mod;
    n/=2;
  }
  return cur;
}

ll div_mod(int a){
  return pow_mod(a,mod-2);
}
// factを前計算
void calcFact(){
  fact[0]=1;
  for(int i=1;i<MAX_FACT;i++)
    fact[i]=(fact[i-1]*i)%mod;
}
// aCbのmodを計算
// 計算量はO(log(fact[a])+log(fact[a-b]))程度
ll comb(ll a,ll b){
  if(a<0||b<0)return 0;
  else if(b==0)return 1;
  ll res=fact[a];
  res=(res*div_mod(fact[b]))%mod;
  res=(res*div_mod(fact[a-b]))%mod;
  return res;
}

int main(){
  calcFact();
  cin>>H>>W>>sy>>sx>>gy>>gx;
  vector<pii> cand;
  ll res=0;
  cand.push_back(pii(gy,gx));
  // left
  cand.push_back(pii(gy,-(W-gx)));
  // right
  cand.push_back(pii(gy,W+gx));
  // up
  cand.push_back(pii(-(H-gy),gx));
  // down
  cand.push_back(pii(H+gy,gx));
  // left up
  cand.push_back(pii(-(H-gy),-(W-gx)));
  // right up
  cand.push_back(pii(-(H-gy),W+gx));
  // left down
  cand.push_back(pii(H+gy,-(W-gx)));
  // right down
  cand.push_back(pii(H+gy,W+gx));
  int minDist=1<<30;
  for(int i=0;i<9;i++)
    minDist=min(minDist,abs(sy-cand[i].first)
                +abs(sx-cand[i].second));
  for(int i=0;i<9;i++){
    int a=abs(sy-cand[i].first);
    int b=abs(sx-cand[i].second);
    if(minDist==(a+b))
      res=(res+comb(a+b,a))%mod;
  }
  cout<<res<<endl;
  
  return 0;
}