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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

r1, c1, r2, c2 = map(int, input().split())

M = r2+c2+10#1000010
mod = 10**9+7

f    = [1]*M
finv = [1]*M
p = 1
for i in range(1, M):
    p = p*i%mod
    f[i] = p
    finv[i] = pow(p, mod-2, mod)

def comb(n, k):
    return f[n] * finv[n-k] * finv[k] % mod

def count(R,C):
    ret = ((1<<R+C+1)-1) % mod

    for c in range(C):
        x1 = comb(R+c, c)
        x2 = ((1<<C-c)-1) % mod
        ret = (ret-x1*x2) % mod

    for r in range(R):
        x1 = comb(C+r, r)
        x2 = ((1<<R-r)-1) % mod
        ret = (ret-x1*x2) % mod

    return ret

ans = count(r2, c2) + count(r1-1, c1-1) - count(r2, c1-1) - count(r1-1, c2)
ans %= mod
print(ans)