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# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
class Program
{
static long factMax = 1000010;//nCkにおけるnの最大値
static long[] factorials = new long[factMax+1];//i!(mod p)をO(n)で先にメモ。
static long[] factorialRs = new long[factMax+1];//i!^-1(mod p), pは素数でO(n)。
static long p = 1000000007;

	static void Main()
	{
    factorials[0] = 1;
		factorialRs[factMax] = DivideModFactorial(factMax,p);
		for(long i = 1; i <= factMax; i++)
		{
		  factorials[i] = (factorials[i-1]*i)%p;//i!(mod p)
		  factorialRs[factMax-i] = (factorialRs[factMax+1-i]*(factMax+1-i))%p;//逆元も先にメモ
		}

    string[] input = Console.ReadLine().Split(' ');
		long r1 = long.Parse(input[0]);
		long c1 = long.Parse(input[1]);
  	long r2 = long.Parse(input[2]);
    long c2 = long.Parse(input[3]);

    long answer = 0;
    for(long i = r1; i <= r2; i++)
    {
      answer += Comb(i+c2+1,i+1,p);
      answer %= p;
      if(answer < 0) answer += p;

      answer -= Comb(i+c1,i+1,p);
      answer %= p;
      if(answer < 0) answer += p;
    }

    Console.WriteLine(answer);
    
	}

  static long MathPow(long x, long a)//戻り値はx^a。O(loga)
  {
    long num = 2;
    long answer = 1;
    long check = a;
    long memo = x;
  
    while(check > 0)
    {
      if(check % num == num / 2)
      {
        check -= num / 2;
        answer *= memo;
      }
    memo *= memo;
    num *= 2;
    }
    return answer;
  }

  static long DivideMod(long x, long a, long p)//戻り値はx^a(mod p)。O(loga)
  {
    long num = 2;
    long answer = 1;
    long check = a;
    long memo = x%p;
    
    while(check > 0)
    {
      if(check % num == num / 2)
      {
        check -= num / 2;
        answer *= memo;
        answer %= p;
      }
    memo *= memo;
    memo %= p;
    num *= 2;
    }
    return answer;
  }

  static long DivideModReverse(long x, long p)//戻り値はx^-1(mod p), pは素数。O(logp)
  {
    long answer = DivideMod(x, p-2, p);
    return answer;
  }

  static long ModFactorial(long x, long p)//戻り値はx!(mod p)
  {
    long answer = 1;
    for(int i = 2; i <= x; i++)
    {
      answer *= i;
      answer %= p;
    }
    return answer;
  }

  static long DivideModFactorial(long x, long p)//戻り値はx!^-1(mod p), pは素数
  {
    long answer = 1;
    for(long i = x; i >= 2; i--)
    {
      answer *= DivideModReverse(i, p);
      answer %= p;
    }
    return answer;
  }

  static long Comb(long a, long b, long p)//組み合わせC(a,b)(mod p)をO(1)で返す
  {
    if(a-b < 0) return 0;//引数a-bが負なら0を返す
    long answer = 1;
    answer *= factorials[a];
    answer %= p;
    answer *= factorialRs[a-b];
    answer %= p;
    answer *= factorialRs[b];
    answer %= p;
    return answer;
  }
  
}