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;
using System.Collections.Generic;
using static System.Console;
using static ConsoleApp.Util;
using static System.Math;
using System.Numerics;

namespace ConsoleApp
{
    class Program
    {
       
        static void Main(string[] args)
        {
            var arr = ReadLine().Split().Select(long.Parse).ToArray();
            long r1 = arr[0];
            long c1 = arr[1];
            long r2 = arr[2];
            long c2 = arr[3];

            //f(1,1)からf(i,j)までの経路総数
            long[,] dp = new long[r2 + 10,c2 + 10];
            dp[0, 0] = 1;

            long t = (1000000000 + 7);

            for (int i = 0; i <= r2;i++)
            {
                for (int j = 0; j <= c2;j++)
                {
                    //x軸＋１
                    dp[i + 1, j] = (dp[i, j]  + dp[i+1,j])%t;

                    //y軸+1
                    dp[i, j + 1] = (dp[i, j] + dp[i,j+1]) % t;
                   

             
                }
            }

            long sum = 0;
            for (long i = r1; i <= r2;i++)
            {
                for (long j = c1; j <= c2;j++)
                {
                    sum =(sum + dp[i, j] )% t;
                }
            }

            WriteLine(sum);
        }

        
    }


    static class Util{
        public static void ChMin(ref long a,long b){
            
            if(a > b){
                a = b;
            }
        }

        public static void ChMax(ref long a,long b){
            if(a < b){
                a = b;
            }
        }



		
    }
}