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 int N=3030,mod=998244353;
int A,B,C,D,f[N][N],inv[N];
int main(){
    scanf("%d%d%d%d",&A,&B,&C,&D);
    inv[1]=1;
    for(int i=2;i<N;i++)inv[i]=mod-1ll*mod/i*inv[mod%i]%mod;
    f[A][B]=1;
    for(int i=B+1;i<=D;i++)
        f[A][i]=1ll*f[A][i-1]*A%mod;
    for(int i=A+1;i<=C;i++)
        f[i][B]=1ll*f[i-1][B]*B%mod;
    for(int i=A+1;i<=C;i++)
        for(int j=B+1,dat=0,t1=1,t2=1;j<=D;j++)
            dat=(dat+1ll*t1*(j-1)%mod*f[i-1][j-1])%mod,
            f[i][j]=(1ll*dat*t2+1ll*f[i-1][j]*j)%mod,
            t1=1ll*t1*inv[i]%mod,t2=1ll*t2*i%mod;
    cout<<f[C][D];
    return 0;
}