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 long long mo = 998244353;

long long dp[3010][3010];
int a, b, c, d;
int main()
{
   cin>>a>>b>>c>>d;

   c-=a;
   d-=b;

   dp[0][0] = 1;

   for (int i = 0;i<=c;i++)
    for (int j = 0;j<=d;j++)
    if (i+j)
   {
       if (i) dp[i][j]+=dp[i-1][j]*(j + b);
       if (j) dp[i][j]+=dp[i][j-1]*(i + a);
       if (i && j)
       dp[i][j]-=dp[i-1][j-1] * (j - 1 + b) * ( i - 1 + a);
       dp[i][j]%=mo;
       if (dp[i][j] < 0) dp[i][j]+=mo;
   }

   cout<<dp[c][d];
}