Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.

Let us define a function f(r, c) as follows:

* f(r,c) :=  (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)



Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).

Constraints

* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


r_1 c_1 r_2 c_2


Output

Print the sum of f(i, j) modulo (10^9+7).

Examples

Input

1 1 2 2


Output

14


Input

314 159 2653 589


Output

602215194
#include<bits/stdc++.h>
using namespace std;
int64_t path(vector<int64_t> f,vector<int64_t> fi,int p,int r,int c){
  return f.at(r+c)*fi.at(r)%p*fi.at(c)%p;
}
int main(){
  int r1,c1,r2,c2;
  cin>>r1>>c1>>r2>>c2;
  int p=1000000007;
  int n=r2+c2+2;
  vector<int64_t> f(n+1,1);
  for(int i=1;i<=n;i++)
    f.at(i)=f.at(i-1)*i%p;
  vector<int64_t> fi(n+1,1);
  int q=p-2;
  vector<bool> b;
  while(q>0){
    b.push_back(q%2==1);
    q/=2;
  }
  for(int d=(int)b.size()-1;d>=0;d--){
    fi.at(n)=fi.at(n)*fi.at(n)%p;
    if(b.at(d))
      fi.at(n)=fi.at(n)*f.at(n)%p;
  }
  for(int i=n;i>=1;i--)
    fi.at(i-1)=fi.at(i)*i%p;
  cout<<(path(f,fi,p,r2+1,c2+1)-path(f,fi,p,r2+1,c1)-path(f,fi,p,r1,c2+1)+path(f,fi,p,r1,c1)+p*2)%p<<endl;
}