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 <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define LL long long
#define rint register int
#define MAXN 3005
#define P 998244353
#define INF 0x3f3f3f3f
using namespace std;

int a, b, c, d, f[MAXN][MAXN];

int main()
{
    scanf("%d%d%d%d", &a, &b, &c, &d);
    f[a][b]=1;
    for(rint i=1; i<=c; ++i)
        for(rint j=1; j<=d; ++j)
        {
            f[i][j]=(f[i][j]+1LL*f[i-1][j]*j+1LL*f[i][j-1]*i)%P;
            f[i][j]=(f[i][j]-1LL*(i-1)*(j-1)*f[i-1][j-1]%P+P)%P;
        }
    printf("%d\n", f[c][d]);
    return 0;
}