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;

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  int A, B, C, D;
  cin >> A >> B >> C >> D;
  const int MOD = 998244353;
  vector<vector<int>> dp(C + 2, vector<int>(D + 2, 0));
  dp[C][D] = 1;
  for (int i = C; i >= A; i--) {
    for (int j = D; j >= B; j--) {
      dp[i][j - 1] += 1ll * i * dp[i][j] % MOD;
      dp[i][j - 1] %= MOD;
      dp[i - 1][j] += 1ll * j * dp[i][j] % MOD;
      dp[i - 1][j] %= MOD;
      dp[i - 1][j - 1] -= 1ll * (i - 1) * (j - 1) * dp[i][j] % MOD;
    }
  }
  cout << (dp[A][B] + MOD) % MOD << "\n";
  return 0;
}