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 int long long
#define double long double
using namespace std;
const int MOD = 998244353;
const int INF = 1e12;
using Graph = vector<vector<int>>;

int dp[4000][4000];

signed main(){
  int A, B, C, D;
  cin >> A >> B >> C >> D;
  dp[A][B] = 1;
  for( int h = A; h <= C; h++ ){
    for( int w = B; w <= D; w++ ){
      dp[h][w] += dp[h-1][w]*w;
      dp[h][w] %= MOD;
      dp[h][w] += dp[h][w-1]*h;
      dp[h][w] %= MOD;
      //重複分を引く
      int hiku = (dp[h-1][w-1]*(h-1)%MOD)*(w-1);
      hiku %= MOD;
      dp[h][w] -= hiku;
      while( dp[h][w] < 0 ) dp[h][w] += MOD;
    }
  }

  cout << dp[C][D] << endl;

}