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 <cassert>
#include <iostream>
using namespace std;
const int M = 998244353;

int main() {
    int A, B, C, D; cin >> A >> B >> C >> D;
    long long dp[C+1][D+1];
    dp[A][B] = 1;
    for (int h = A+1; h <= C; h++) dp[h][B] = B * dp[h-1][B] % M;
    for (int w = B+1; w <= D; w++) dp[A][w] = A * dp[A][w-1] % M;
    for (int h = A + 1; h <= C; h++)
        for (int w = B + 1; w <= D; w++)
            dp[h][w] = (h * dp[h][w-1] + w * dp[h-1][w] - (h - 1) * (w - 1) * dp[h-1][w-1]) % M;

    cout << (dp[C][D] % M + M) % M << endl;
}