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;
#define int long long

const int N=3009,mod=998244353;
int dp[N][N],a,b,c,d;

int fn(int i,int j){
    if(i==a and j==b)return 1;
    if(dp[i][j]!=-1)return dp[i][j];
    dp[i][j]=0;
    if(i>a)dp[i][j]=(dp[i][j]+j*fn(i-1,j)%mod)%mod;
    if(j>b)dp[i][j]=(dp[i][j]+i*fn(i,j-1)%mod)%mod;;
    if(i>a and j>b)dp[i][j]+=mod-(i-1)*(j-1)%mod*fn(i-1,j-1)%mod;
    dp[i][j]%=mod;
    return dp[i][j];
}
int32_t main(){
    ios::sync_with_stdio(0),cin.tie(0);
    memset(dp,-1,sizeof dp);
    cin >> a >> b >> c >> d;
    cout << fn(c,d);
}