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 CompLib.Util;
using System;

public class Program
{
    int R1, R2, C1, C2;

    ModInt[] Fact;
    public void Solve()
    {
        var sc = new Scanner();
        R1 = sc.NextInt();
        C1 = sc.NextInt();
        R2 = sc.NextInt();
        C2 = sc.NextInt();


        Fact = new ModInt[500001];
        Fact[0] = 1;
        for (int i = 1; i <= 500000; i++)
        {
            Fact[i] = Fact[i - 1] * i;
        }
        // f(r,c) = (r+c) C r

        // rを固定
        ModInt ans = 0;
        for(int i = R1; i <= R2; i++)
        {
            ans += Calc(i);
        }

        Console.WriteLine(ans);
    }

    private ModInt Calc(int r)
    {
        return Calc2(r, C2) - Calc2(r, C1 - 1);
    }

    // Σ (r+c) C c
    private ModInt Calc2(int r, int c)
    {
        // r = 0 return = c + 1
        // r = 1 return 1 + 2 + 3 + 4 ...

        // パスカルの三角形の r+1,c

        return C(r + 1 + c, c);

    }

    private ModInt C(int r, int c)
    {
        return Fact[r] * ModInt.Inverse(Fact[r - c] * Fact[c]);
    }

    public static void Main(string[] args) => new Program().Solve();
}


#region ModInt
/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt
{
    /// <summary>
    /// 剰余を取る値．
    /// </summary>
    public const long Mod = (int)1e9 + 7;

    /// <summary>
    /// 実際の数値．
    /// </summary>
    public long num;
    /// <summary>
    /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
    /// </summary>
    /// <param name="n">インスタンスが持つ値</param>
    /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
    public ModInt(long n) { num = n; }
    /// <summary>
    /// このインスタンスの数値を文字列に変換します．
    /// </summary>
    /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
    public override string ToString() { return num.ToString(); }
    public static ModInt operator +(ModInt l, ModInt r) { l.num += r.num; if (l.num >= Mod) l.num -= Mod; return l; }
    public static ModInt operator -(ModInt l, ModInt r) { l.num -= r.num; if (l.num < 0) l.num += Mod; return l; }
    public static ModInt operator *(ModInt l, ModInt r) { return new ModInt(l.num * r.num % Mod); }
    public static implicit operator ModInt(long n) { n %= Mod; if (n < 0) n += Mod; return new ModInt(n); }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(ModInt v, long k) { return Pow(v.num, k); }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(long v, long k)
    {
        long ret = 1;
        for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
            if ((k & 1) == 1) ret = ret * v % Mod;
        return new ModInt(ret);
    }
    /// <summary>
    /// 与えられた数の逆元を計算します．
    /// </summary>
    /// <param name="v">逆元を取る対象となる数</param>
    /// <returns>逆元となるような値</returns>
    /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
    public static ModInt Inverse(ModInt v) { return Pow(v, Mod - 2); }
}
#endregion

namespace CompLib.Util
{
    using System;
    using System.Linq;
    class Scanner
    {
        private int _index;
        private string[] _line;
        const char separator = ' ';

        public Scanner()
        {
            _index = 0;
            _line = new string[0];
        }

        public string Next()
        {
            while (_index >= _line.Length)
            {
                _line = Console.ReadLine().Split(separator);
                _index = 0;
            }
            return _line[_index++];
        }
        public int NextInt() => int.Parse(Next());
        public long NextLong() => long.Parse(Next());
        public double NextDouble() => double.Parse(Next());
        public char NextChar() => Next()[0];
        public char[] NextCharArray() => Next().ToCharArray();

        public string[] Array()
        {
            _line = Console.ReadLine().Split(separator);
            _index = _line.Length;
            return _line;
        }
        public int[] IntArray() => Array().Select(int.Parse).ToArray();
        public long[] LongArray() => Array().Select(long.Parse).ToArray();
        public double[] DoubleArray() => Array().Select(double.Parse).ToArray();
    }
}