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 const p=998244353;
int f[3005][3005],g[3005][3005];
int pw(int x,int y)
{
    int res=1;
    while(y)
    {
        if(y&1)res=1ll*res*x%p;
        x=1ll*x*x%p;
        y>>=1;
    }
    return res;
}
int main()
{
    int a,b,c,d;
    scanf("%d%d%d%d",&a,&b,&c,&d);
    f[a][b]=1;
    for(int i=a+1;i<=c;i++)
    {
        for(int j=b;j<=d;j++)
        {
            g[i][j]=1ll*(g[i][j-1]+f[i-1][j-1])*(i-1)%p;
            f[i][j]=(1ll*f[i-1][j]*j+g[i][j])%p;
        }
    }
    int ans=0;
    for(int i=b;i<=d;i++)ans=(ans+1ll*f[c][i]*pw(c,d-i))%p;
    printf("%d",ans);
    return 0;
}