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<bits/stdc++.h>
using namespace std;
template <typename T> using vec = vector<T>;

const long MOD = 998244353;

int main()
{
    long a,b,c,d;
    cin >> a >> b >> c >> d;
    long dp[c+1][d+1];
    dp[a][b] = 1;
    for (long i = a; i < c; i++)
    {
        dp[i+1][b] = dp[i][b] % MOD * b % MOD;
    }
    for (long j = b; j < d; j++)
    {
        dp[a][j+1] = dp[a][j] % MOD * a % MOD;
    }

    for (long i = a; i < c; i++)
    {
        for (long j = b; j < d; j++)
        {
            dp[i+1][j+1] = ( (dp[i][j+1] * (j+1) + dp[i+1][j] * (i+1))%MOD - (dp[i][j] * (i*j))%MOD + MOD ) % MOD;
        }
    }
    cout << dp[c][d] << endl;
}