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)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using uint = unsigned int;
template <typename T>
using Graph = vector<vector<T>>;
const int MOD = 998244353;
const ld PI = acosl(-1);

struct COM {
    vector<ll> inv, fac, finv;
    COM(int N) : inv(N), fac(N), finv(N) {
        inv[1] = 1;
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        for (int i = 2; i < N; ++i) {
            inv[i] = MOD - (MOD / i) * inv[MOD % i] % MOD;
            fac[i] = fac[i - 1] * i % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    ll calc(int a, int b) {
        return (fac[a] * finv[b] % MOD) * finv[a - b] % MOD;
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int A, B, C, D;
    cin >> A >> B >> C >> D;

    COM com(C * D + 10);
    int m = C * D - A * B, n = C + D - A - B;
    ll ans = com.calc(m, n);

    vector<vector<ll>> dp(C + 3, vector<ll>(D + 3));
    dp[A][B] = 1;
    for (int i = A; i <= C; ++i) {
        for (int j = B; j <= D; ++j) {
            (dp[i + 1][j] += (dp[i][j] * j) % MOD) %= MOD;
            (dp[i][j + 1] += (dp[i][j] * i) % MOD) %= MOD;
            (dp[i + 1][j + 1] -= (dp[i][j] * i % MOD) * j % MOD) %= MOD;
        }
    }

    if (dp[C][D] < 0)
        dp[C][D] += MOD;
    cout << dp[C][D] << endl;

    return 0;
}