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
#dt = {} for i in x: dt[i] = dt.get(i,0)+1
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):
    # n C r = n!*inverse(r!)*inverse((n-r)!) 
    #ans = ((fact[N] * factorialNumInverse[R])% p * factorialNumInverse[N-R])% p
    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)
#dt = {} for i in x: dt[i] = dt.get(i,0)+1
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):
    # n C r = n!*inverse(r!)*inverse((n-r)!) 
    #ans = ((fact[N] * factorialNumInverse[R])% p * factorialNumInverse[N-R])% p
    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)