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;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
//const ll mod=1e9+7;
const ll mod=998244353;
const ll inf=1LL<<61;

int main() {
  ll h,w,y,x;
  cin >> h >> w >> y >> x;
  mat dp(y+1,vec(x+1));
  dp[h][w]=1;
  for(ll i=h;i<=y;i++) {
    for(ll j=w;j<=x;j++) {
      if(i-h|j-w) {
        dp[i][j]=(dp[i-1][j]*j+dp[i][j-1]*i-dp[i-1][j-1]*(i-1)*(j-1)%mod+mod)%mod;
      }
    }
  }
  cout << dp[y][x] << endl;
}