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<iostream>
#define mod 998244353
using namespace std;
int a, b, c, d, i, j;
int dp[3005][3005], e[3005][3005];
int main(){
    cin>> a >> b >> c >> d;
    dp[c][d] = 1;
    for(i = c; i > a; i--){
        for(j = d; j > b; j--){
            e[i][j] = (dp[i][j] + e[i][j + 1] * 1LL * i) % mod;
        }
        for(j = d; j >= b; j--){
            dp[i][j] = (dp[i][j] + e[i][j + 1]) % mod;
            dp[i - 1][j] = (dp[i - 1][j] + dp[i][j] * 1LL * j) % mod;
        }
    }
    for(j = d; j > b; j--){
        dp[a][j - 1] = (dp[a][j - 1] + dp[a][j] * 1LL * a) % mod;
    }
    cout<< dp[a][b];
}