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 CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>

int const nmax = 3000;
int const modulo = 998244353;

int dp[5 + nmax][5 + nmax];

int main() {
  int n, m, q, p;
  std::cin >> n >> m >> p >> q;
  dp[p][q] = 1;
  for(int i = p; n <= i; i--)
    for(int j = q; m <= j; j--)
      dp[i][j] += (modulo + 1LL * dp[i][j + 1] * i + 1LL * dp[i + 1][j] * j - 1LL * dp[i + 1][j + 1] * i % modulo * j % modulo) % modulo;
  std::cout << dp[n][m];
}