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>
#define meow(args...) fprintf(stderr, args)
typedef unsigned u32;
typedef unsigned long long u64;
const u32 P=998244353;
u32 f[3001][3001], g[3001][3001];
int main() {
	int a, b, c, d;
	scanf("%d%d%d%d", &a, &b, &c, &d);
	f[a][b]=g[a][b]=1;
	for(int i=b+1; i<=d; ++i) f[a][i]=1llu*f[a][i-1]*a%P;
	for(int i=a+1; i<=c; ++i) g[i][b]=1llu*g[i-1][b]*b%P;
	g[a][b]=0;
	for(int i=a+1; i<=c; ++i) {
		for(int j=b+1; j<=d; ++j) {
			f[i][j]=(1llu*i*(f[i][j-1]+g[i][j-1]))%P;
			g[i][j]=(1llu*j*g[i-1][j]+f[i-1][j])%P;
		}
	}
	printf("%u\n", (f[c][d]+g[c][d])%P);
	return 0;
}