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: Accepted)

#pragma GCC optimize("O2")
#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0)
//#define MULTI_TEST
//#define GOOGLE
#ifdef LOCAL
#include "debugger.h"
#else
#define db(...)
#endif
#define ll long long
#define v32 vector<int>
#define v64 vector<ll>
#define s32 set<int>
#define s64 set<ll>
#define p32 pair<int, int>
#define p64 pair<ll, ll>
#define sz(v) (ll)(v).size()
#define fi first
#define se second
#define ln '\n'
const ll MOD = 998244353;
inline ll add(ll a, ll b, ll m) { return (((a % m + b % m) % m + m) % m); }
inline ll mul(ll a, ll b, ll m) { return (((a % m * b % m) % m + m) % m); }

const int N = 3005;
int dp[N][N];
int go(int i, int j, int a, int b)
{
	if (i == a && j == b)
		return 1;
	if (dp[i][j] != -1)
		return dp[i][j];
	int ans = 0;
	if (i - 1 >= a)
	{
		int tmp = mul(j, go(i - 1, j, a, b), MOD);
		ans = add(ans, tmp, MOD);
	}
	if (j - 1 >= b)
	{
		int tmp = mul(i, go(i, j - 1, a, b), MOD);
		ans = add(ans, tmp, MOD);
	}
	if (i - 1 >= a && j - 1 >= b)
	{
		int tmp = mul(i - 1, j - 1, MOD);
		tmp = mul(tmp, go(i - 1, j - 1, a, b), MOD);
		ans = add(ans, MOD - tmp, MOD);
	}
	return dp[i][j] = ans;
}
void solve(int tc)
{
	int a, b, c, d;
	cin >> a >> b >> c >> d;
	memset(dp, -1, sizeof(dp));
	cout << go(c, d, a, b) << ln;
	return;
}

int main()
{
#ifndef LOCAL
	FASTIO;
#endif
	int t = 1;
#ifdef LOCAL
	clock_t start, stop;
	start = clock();
#endif
#ifdef MULTI_TEST
	cin >> t;
#endif
	for (int tc = 1; tc < t + 1; tc++)
	{
#ifdef GOOGLE
		cout << "Case #" << tc << ": ";
#endif
		solve(tc);
	}
#ifdef LOCAL
	stop = clock();
	cerr << "Time :" << fixed << setprecision(5) << double(stop - start) / double(CLOCKS_PER_SEC) << "s" << ln;
#endif
	return 0;
}