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>

#define MOD 998244353LL

using namespace std;

int main() {
  int A, B, C, D;
  cin >> A;
  cin >> B;
  cin >> C;
  cin >> D;
  vector <vector <long long> > dp(C + 1, vector <long long> (D + 1, 0LL));
  dp[A][B] = 1LL;
  for (int c = A; c <= C; ++c) {
    for (int d = B; d <= D; ++d) {
      if (c == A && d == B) continue;
      dp[c][d] = 
        ((long long)d * dp[c - 1][d] % MOD)
        + ((long long)c * dp[c][d - 1] % MOD)
        - (dp[c - 1][d - 1] * (long long)(c - 1) * (d - 1) % MOD);
      if (dp[c][d] < 0) dp[c][d] += MOD;
      dp[c][d] %= MOD;
    }
  }

  cout << dp[C][D] << endl;
  return 0;
}