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;

const int N = 3003, md = 998244353;
int dp[N][N];

inline int add(int x, int y) {
  x += y;
  if (x >= md) {
    x -= md;
  }
  if (x < 0) {
    x += md;
  }
  return x;
}

inline int mult(int x, int y) {
  return (long long) x * y % md;
}

int main() {
  int a, b, c, d;
  cin >> a >> b >> c >> d;
  dp[a][b] = 1;
  for (int i = a; i <= c; i++) {
    for (int j = b + (i == a); j <= d; j++) {
      dp[i][j] = add(mult(dp[i - 1][j], j), mult(dp[i][j - 1], i));
      dp[i][j] = add(dp[i][j], -mult(dp[i - 1][j - 1], mult(i - 1, j - 1)));
    }
  }
  cout << dp[c][d] << '\n';
  return 0;
}