p02782 AtCoder Beginner Contest 154 - Many Many Paths
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).

Examples

Input

1 1 2 2


Output

14


Input

314 159 2653 589


Output

602215194
```python
r1, c1, r2, c2 = map(int, input().split())
mod = int(1e+9) + 7
def inved(x):
  a, b, c, d, k, l = 1, 0, 0, 1, x, mod
  while l != 0:
    a, b, c, d = c, d, a - c * (k // l), b - d * (k // l)
    k, l = l, k % l
  return a % mod
frac = [1]
for i in range(r2 + c2 + 2):
  frac.append(((i+1) * frac[i]) % mod)
fracr2c2 = frac[r2 + c2 + 2]
fracr1c2 = frac[r1 + c2 + 1]
fracr2c1 = frac[r2 + c1 + 1]
fracr1c1 = frac[r1 + c1]
fracr2 = frac[r2 + 1]
fracr1 = frac[r1]
fracc2 = frac[c2 + 1]
fracc1 = frac[c1]
g = 0
g += (fracr2c2 * inved(fracr2) * inved(fracc2)) % mod
g %= mod
g += (fracr1c1 * inved(fracr1) * inved(fracc1)) % mod
g %= mod
g -= (fracr1c2 * inved(fracr1) * inved(fracc2)) % mod
g %= mod
g -= (fracr2c1 * inved(fracr2) * inved(fracc1)) % mod
g %= mod
print(g)

```