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>
#define F first
#define S second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pi;
const int maxn=3005;
const int M=998244353;
int dp[maxn][maxn],A,B,C,D;
void add(int &x,int y){
    x+=y; if(x>=M) x-=M;
}
int main(){
    cin >> A >> B >> C >> D;
    dp[A][B]=1;
    for (int i=A;i<=C;i++)
    for (int j=B;j<=D;j++){
        add(dp[i][j],1ll*dp[i-1][j]*j%M);
        add(dp[i][j],1ll*dp[i][j-1]*i%M);
        add(dp[i][j],M-1ll*dp[i-1][j-1]*(i-1)*(j-1)%M);
    }
    cout << dp[C][D] << endl;
    return 0;
}