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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.io.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };

	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int r1 = fs.nextInt(),c1 = fs.nextInt();
		int r2 = fs.nextInt(),c2 = fs.nextInt();
		pre_factrials((int)1e7);
		long ans = 0;
		for(int i=c1;i<=c2;i++) {
			long s = fastFermatComb(r2+i+1, i+1);
			long t = fastFermatComb(r1-1+i+1, i+1);
			long st = sub(s,t);
			ans = plus(ans,st);
		}
		System.out.println(ans);
	}
	//Fast Fermat Combination
	static long factorials[];

	static void pre_factrials(int n) {
		factorials = new long[n + 1];
		factorials[0] = 1;
		for (int i = 0; i < n; i++) {
			factorials[i + 1] = mul(i + 1, factorials[i]);
		}
	}

	static long fastFermatComb(long a, long b) {
		if (factorials.length == 0)
			System.err.println("error : factorials has not been culculated!! do [pre_factorial] method");
		long af = factorials[(int) a];
		long bf = factorials[(int) b];
		long abf = factorials[(int) (a - b)];
		long res = div(af,mul(bf,abf));
		return res;
	}
	
	//MOD culculations
	static long plus(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x + y) % mod;
		return res;
	}

	static long sub(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x - y + mod) % mod;
		return res;
	}

	static long mul(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * y % mod;
		return res;
	}

	static long div(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * modpow(y, mod - 2) % mod;
		return res;
	}

	//modの元での高速pow()
	static long modpow(long a, long p) {
		if (p == 0)
			return 1;
		if (p % 2 == 0) {
			long halfP = p / 2;
			long root = modpow(a, halfP);
			return root * root % mod;
		} else
			return a * modpow(a, p - 1) % mod;
	}


}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}