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;

const int M = 998244353, N = 3005;
long long dp[N][N];

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

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

    dp[A][B] = 1;

    for (int i=A+1; i<=C; i++)  dp[i][B] = (B*dp[i-1][B])%M;
    for (int i=B+1; i<=D; i++)  dp[A][i] = (A*dp[A][i-1])%M;

    for (int i=A+1; i<=C; i++) {
        for (int j=B+1; j<=D; j++) {
            dp[i][j] = (dp[i-1][j]*j+dp[i][j-1]*i)%M;
            long long ww = (1LL*(i-1)*(j-1)*dp[i-1][j-1])%M;
            dp[i][j] = (dp[i][j]+M-ww)%M;
        }
    }
    cout<<dp[C][D]<<endl;
}