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:
// B.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 3030, mod = 998244353;

int n, m, a, b, dp[MAX_N][MAX_N];

int main()
{
    scanf("%d%d%d%d", &a, &b, &n, &m);
    dp[a][b] = 1;
    for (int i = a; i <= n; i++)
        for (int j = b; j <= m; j++)
            if (i != a || j != b)
                dp[i][j] = (0LL + 1LL * dp[i - 1][j] * j % mod + 1LL * dp[i][j - 1] * i % mod + mod - 1LL * dp[i - 1][j - 1] * (i - 1) % mod * (j - 1) % mod) % mod;
    printf("%d\n", dp[n][m]);
    return 0;
}