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;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
ll a,b,c,d,mod=998244353,dp[3005][3005];
int main(){
	a=read(),b=read(),c=read(),d=read();
    dp[a][b]=1;
    for(int i=a;i<=c;i++)
        for(int j=b;j<=d;j++){
        	if(i+1<=c)dp[i+1][j]=(dp[i+1][j]+dp[i][j]*j)%mod;
        	if(j+1<=d)dp[i][j+1]=(dp[i][j+1]+dp[i][j]*i)%mod;
        	if(i+1<=c and j+1<=d)dp[i+1][j+1]=(dp[i+1][j+1]+mod-dp[i][j]*i*j%mod)%mod;
        }
    cout << dp[c][d];
	return 0;
}