Problem: 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.

Examples

Input

1 1 2 2


Output

3


Input

2 1 3 4


Output

65


Input

31 41 59 265


Output

387222020

Solution (in PYTHON3):

Solution:
a, b, c, d = map(int, input().split())
mod = 998244353

dp1 = [[0 for j in range(d)] for i in range(c)] # 一番下の行に黒一つは下に追加で得られた場合に限って良い (下に追加する前の盤を初期状態から作れるので)
dp2 = [[0 for j in range(d)] for i in range(c)] # その状態が右に追加することでしか得られない場合 (一番下の行に黒複数 or a==c)
dp2[a-1][b-1] = 1

for i in range(a-1, c):
    for j in range(b-1, d):
        if i==a-1 and j==b-1: continue
        if i-1 >= 0:
            dp1[i][j] = (j+1)*(dp1[i-1][j]+dp2[i-1][j])
            dp1[i][j] %= mod
        if j-1 >= 0:
            dp2[i][j] = dp1[i][j-1] + (i+1)*dp2[i][j-1]
            dp2[i][j] %= mod

ans = (dp1[c-1][d-1] + dp2[c-1][d-1]) % mod
print(ans)