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 ll = int_fast64_t;
#define REP(i,b,e) for(ll i=b; i<e; i++)
const int MOD = 998244353;

int main(){
	int a, b, c, d;
	scanf("%d %d %d %d", &a, &b, &c, &d);
	
	ll dp[c+1][d+1];
	dp[a][b] = 1;
	REP(j, b+1, d+1) dp[a][j] = dp[a][j-1] * a % MOD;
	REP(i, a+1, c+1) dp[i][b] = dp[i-1][b] * b % MOD;

	REP(i, a+1, c+1){
		REP(j, b+1, d+1){
			dp[i][j] = (dp[i-1][j]*j%MOD + dp[i][j-1]*i%MOD) % MOD;
			dp[i][j] -= dp[i-1][j-1]*(i-1)%MOD * (j-1)%MOD;
			if(dp[i][j]<0) dp[i][j] += MOD;
		}
	}

	printf("%ld\n", dp[c][d]);
	return 0;
}