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() {
 const long long mod = 998244353;
 int a, b, c, d; cin >> a >> b >> c >> d;

 vector<vector<long long>> dp(c + 1, vector<long long>(d + 1, 0));
 dp[a][b] = 1;


 for (int x = a; x < c + 1; x++) {
  for (int y = b; y < d + 1; y++) {
   if (x != a or y != b) {
    dp[x][y] = dp[x - 1][y] * y + dp[x][y - 1] * x - dp[x - 1][y - 1] * (x - 1) * (y - 1);
    dp[x][y] = dp[x][y]%mod;
    dp[x][y]+=2*mod;
   }
  }
 }
 dp[c][d]+=mod;
  dp[c][d]%=mod;
 cout << dp[c][d]<< endl;
}