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>
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 4e3+5;
const long long MOD = 998244353;
long long dp[MAXN][MAXN];
int main() {
    long long a,b,c,d;
    cin>>a>>b>>c>>d;
    dp[a][b] = 1;
    
    for(long long i=a;i<=c;i++){
        for(long long j=b;j<=d;j++){
            if(i==a && j==b){
                continue;
            }
           dp[i][j] = ((dp[i][j-1]+dp[i-1][j])%MOD+(1LL*dp[i][j-1]*(i-1))%MOD+(1LL*dp[i-1][j]*(j-1))%MOD)%MOD-((1LL*dp[i-1][j-1]*(i-1)*(j-1))%MOD)+MOD;
           dp[i][j]%=MOD;
        }
    }
    //dp[a][b] = 0;
    cout<<dp[c][d]<<endl;
}