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>
using ll = long long;
using ull = unsigned long long;
const static int MOD = 998244353;
ull dp[3001][3001];

int main(){
    ull A, B, C, D;
    scanf("%llu %llu %llu %llu", &A, &B, &C, &D);
    dp[A][B] = 1;
    for (unsigned i = A + 1; i <= C; i++){
        dp[i][B] = dp[i-1][B]*B %MOD;
    }
    for (unsigned i = B + 1; i <= D; i++){
        dp[A][i] = dp[A][i-1]*A %MOD;
    }
    for (ull i = A + 1; i <= C; i++){
        for (ull j = B + 1; j <= D; j++){
            dp[i][j] = ((dp[i-1][j]*j %MOD) + (dp[i][j-1]*i %MOD) + MOD - ((i-1)*(j-1)%MOD*dp[i-1][j-1]%MOD)) %MOD;
        }
    }
    printf("%llu\n", dp[C][D]);    
	return 0;
}