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;
const int N = 3000 + 10, mod = 998244353;
ll f[N][N][2];
void add(ll &x,ll y){
    (x += y) %= mod;
}
int main(){
    ios::sync_with_stdio(0);
    int a,b,c,d; cin>>a>>b>>c>>d;
    f[a][b][0] = 1;
    for(int i=a;i<=c;i++){
        for(int j=b;j<=d;j++){
            add(f[i+1][j][0],f[i][j][0]*j);
            add(f[i][j+1][1],f[i][j][0]*i);
            add(f[i+1][j][0],f[i][j][1]);
            add(f[i][j+1][1],f[i][j][1]*i);
        }
    }
    ll ans = 0;
    add(ans,f[c][d][0]);
    add(ans,f[c][d][1]);
    cout<<ans<<endl;
}