Problem:
Score : 600 points

Problem Statement

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).

Sample Input 1

1 1 2 2

Sample Output 1

14

For example, there are two paths from the point (0, 0) to the point (1, 1): (0,0) → (0,1) → (1,1) and (0,0) → (1,0) → (1,1), so f(1,1)=2.

Similarly, f(1,2)=3, f(2,1)=3, and f(2,2)=6. Thus, the sum is 14.

Sample Input 2

314 159 2653 589

Sample Output 2

602215194

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
const ll mod=1e9+7;
const int inf=1<<30;

const int MAX=2000010; //nCrのnのMAX+1
ll fac[MAX],finv[MAX],inv[MAX];

void cominit(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for (int i=2;i<MAX;i++){
    fac[i]=fac[i-1]*i%mod;
    inv[i]=mod-inv[mod%i]*(mod/i)%mod;
    finv[i]=finv[i-1]*inv[i]%mod;
  }
}
ll com(int n,int k){ //1≦k≦n≦10^7
  if (n<k) return 0;
  if (n<0||k<0) return 0;
  return fac[n]*(finv[k]*finv[n-k]%mod)%mod;
}
//Add cominit(); in main function.

int main(){
  cominit();
  ll r1,c1,r2,c2;
  cin >> r1 >> c1 >> r2 >> c2;
  ll ans=com(r1+c1,r1);
  for(int i=r1+1;i<=r2;i++){
    ans+=com(i+c1,c1)*((r2-i+1)*(c2-c1)+1)%mod;
    ans%=mod;
  }
  for(int j=c1+1;j<=c2;j++){
    ans+=com(r1+j,r1)*((r2-r1)*(c2-j+1)+1)%mod;
    ans%=mod;
  }
  cout << ans << endl;
}