Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

1 1 2 2

Sample Output 1

3

Any two of the three squares other than the bottom-left square can be painted black.

Sample Input 2

2 1 3 4

Sample Output 2

65

Sample Input 3

31 41 59 265

Sample Output 3

387222020

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define MOD 998244353
typedef long long ll;
ll modpow(ll a, ll n) {
	if (n == 0)
		return 1;
	if (n == 1)
		return (a % MOD);
	if (n % 2 == 1)
		return (a * modpow(a, n - 1) % MOD);
	ll t = modpow(a, n / 2);
	return (t * t % MOD);
}
int main() {
	int a, b, c, d;
	ll ans = 0;
	scanf("%d%d%d%d", &a, &b, &c, &d);
	if (b - d == 1) {
		int tmp = a;
		a = b;
		b = tmp;
		tmp = c;
		c = d;
		d = tmp;
	}
	for (int i = 0; i <= d - b; i++) {
		ans += modpow(a, d - b - i) * modpow(c, i) % MOD;
		ans %= MOD;
		// printf("%d^%d*%d^%d\n", a, d - b - i, c, i);
	}
	printf("%lld\n", ans);
	return 0;
}