You are given a grid with dimension $n$ x $m$ and two points with coordinates $X(x1,y1)$ and $Y(x2,y2)$ . Your task is to find the number of ways in which one can go from point $A(0, 0)$ to point $B (n, m)$ using the $shortest$ possible path such that the shortest path neither passes through $X$ nor through $Y$. 

Consider the above 4 x 4 grid . Our shortest path can't pass through points (1,3) and (3,3) (marked by yellow dots). One of the possible shortest path is from $A$ to $C$ and then from $C$ to $B$.

-----Input:-----
- First line contains $T$, number of testcases. Then the testcases follow. 
- Each testcase contains of a single line of input, six space separated integers $n, m, x1, y1, x2, y2$. 

-----Output:-----
- For each testcase, output in a single line number of ways modulo $998244353$.

-----Constraints-----
- $1 \leq T \leq 10^5$
- $3 \leq n,m \leq 10^5$
- $1 \leq x1, x2 \leq n - 1$
- $1 \leq y1, y2 \leq m - 1$
- $x1 \leq x2$
- $y1 \leq y2$
- $X$ and $Y$ never coincide.

-----Sample Input:-----
1
3 3 1 1 1 2

-----Sample Output:-----
5

Difficulty: UNKNOWN_DIFFICULTY

Solutions:
import sys
input = sys.stdin.readline
(inp, ip) = (lambda : int(input()), lambda : [int(w) for w in input().split()])
N = 100001
p = 998244353
factorialNumInverse = [0] * (N + 1)
naturalNumInverse = [0] * (N + 1)
fact = [0] * (N + 1)

def InverseofNumber(p):
	naturalNumInverse[0] = naturalNumInverse[1] = 1
	for i in range(2, N + 1):
		naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p // i) % p

def InverseofFactorial(p):
	factorialNumInverse[0] = factorialNumInverse[1] = 1
	for i in range(2, N + 1):
		factorialNumInverse[i] = naturalNumInverse[i] * factorialNumInverse[i - 1] % p

def factorial(p):
	fact[0] = 1
	for i in range(1, N + 1):
		fact[i] = fact[i - 1] * i % p

def f(num, den1, den2):
	ans = fact[num] * factorialNumInverse[den1] % p * factorialNumInverse[den2] % p
	return ans
InverseofNumber(p)
InverseofFactorial(p)
factorial(p)
for _ in range(inp()):
	(n, m, x1, y1, x2, y2) = ip()
	tot = f(m + n, m, n)
	a = f(m - y1 + n - x1, m - y1, n - x1)
	aa = f(x1 + y1, x1, y1)
	b = f(m - y2 + n - x2, m - y2, n - x2)
	bb = f(x2 + y2, x2, y2)
	c = f(y2 - y1 + x2 - x1, y2 - y1, x2 - x1)
	ans = (tot - a * aa - b * bb + c * aa * b) % p
	print(ans)