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;
#define PI 3.141592653589793238462
typedef long long ll;
typedef double db;
ll mod=998244353,p[3005][3005];
int main(){
    ll a,b,c,d;
    cin>>a>>b>>c>>d;
    p[a][b]=1;
    for(ll i=a+1;i<=c;i++){
        p[i][b]=(p[i-1][b]%mod*b%mod)%mod;
    }
    for(ll i=b+1;i<=d;i++){
        p[a][i]=(p[a][i-1]%mod*a%mod)%mod;
    }
    for(ll i=a+1;i<=c;i++){
        for(ll j=1+b;j<=d;j++){
            p[i][j]=((p[i-1][j]%mod*j%mod+p[i][j-1]%mod*i%mod-p[i-1][j-1]%mod*(i-1)%mod*(j-1)%mod)%mod+mod)%mod;
        }
    }
    cout<<(p[c][d]%mod+mod)%mod<<endl;
}