Problem:
Score : 600 points

Problem Statement

We have a grid with A horizontal rows and B vertical columns, with the squares painted white. On this grid, we will repeatedly apply the following operation:

Assume that the grid currently has a horizontal rows and b vertical columns. Choose "vertical" or "horizontal".

If we choose "vertical", insert one row at the top of the grid, resulting in an (a+1) \times b grid.

If we choose "horizontal", insert one column at the right end of the grid, resulting in an a \times (b+1) grid.

Then, paint one of the added squares black, and the other squares white.

Assume the grid eventually has C horizontal rows and D vertical columns. Find the number of ways in which the squares can be painted in the end, modulo 998244353.

Constraints

1 \leq A \leq C \leq 3000

1 \leq B \leq D \leq 3000

A, B, C, and D are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the number of ways in which the squares can be painted in the end, modulo 998244353.

Sample Input 1

1 1 2 2

Sample Output 1

3

Any two of the three squares other than the bottom-left square can be painted black.

Sample Input 2

2 1 3 4

Sample Output 2

65

Sample Input 3

31 41 59 265

Sample Output 3

387222020

Solution in PyPy3 (7.3.0) (judge verdict: Accepted)

a,b,c,d=map(int,input().split())
dp=[]
MOD=998244353
for i in range(c+1):
  dp.append([])
  for j in range(d+1):
    dp[-1].append(0)
dp[a][b]=1
for i in range(a,c+1):
  for j in range(b,d+1):
    if i+1<=c and j+1<=d:
      dp[i+1][j+1]=(dp[i+1][j+1]-dp[i][j]*i*j)%MOD
    if i+1<=c:
      dp[i+1][j]=(dp[i+1][j]+dp[i][j]*j)%MOD
    if j+1<=d:
      dp[i][j+1]=(dp[i][j+1]+dp[i][j]*i)%MOD
print(dp[c][d])