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 <bits/stdc++.h>

using namespace std;
typedef long long ll;


ll P = 998244353;
ll DP[3001][3001] = {0};

int main(){
  int A,B,C,D;
  std::cin >> A >> B >> C >> D;
  DP[A][B] = 1;
  for(int a=A+1; a<=C; a++)
    DP[a][B] = DP[a-1][B] * B % P;
  for(int b=B+1; b<=D; b++)
    DP[A][b] = DP[A][b-1] * A % P;
  for(int a=A+1; a<=C; a++)
    for(int b=B+1; b<=D; b++)
      DP[a][b] = (P + b*DP[a-1][b] + a*DP[a][b-1] - (a-1)*(b-1)*DP[a-1][b-1]%P) % P;

  std::cout << DP[C][D] << std::endl;
  return 0;
}