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 N 3030
#define P 998244353
typedef long long LL;

LL f[N][N];
LL a,b,c,d;

int main(){
    cin>>a>>b>>c>>d;
    f[0][0] = 1;
    int n = c - a, m = d - b;
    for(int i = 0; i <= n; i++){
        for(int j = 0; j <= m; j++){
            if(i == 0 && j == 0) continue;
            f[i][j] = 0;
            if(i) f[i][j] += f[i - 1][j] * (j + b) % P;
            if(j) f[i][j] += f[i][j - 1] * (i + a) % P;
            if(i * j) f[i][j] -= f[i - 1][j - 1] * (a + i - 1) % P * (b + j - 1) % P;
            f[i][j] = (f[i][j] % P + P) % P;
        }
    }
    printf("%lld\n", f[n][m]);
    return 0;

}