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;
using ll = long long;

const ll mod = 998244353;

int a, b;

vector<vector<ll> > dp(3001, vector<ll>(3001, -1));

ll solve(int x, int y) {
  if (dp[x][y] < 0) {
    dp[x][y] = 0;
    if (x > a) dp[x][y] += solve(x-1, y)*y%mod;
    if (y > b) dp[x][y] += solve(x, y-1)*x%mod;
    if (x > a and y > b) dp[x][y] = (dp[x][y] - solve(x-1,y-1)*(x-1)%mod*(y-1))%mod;
    if (dp[x][y] < 0) dp[x][y] += mod;
    
  }
  return dp[x][y];
}

int main() {
  int c , d;
  cin >> a >> b >> c >> d;
  dp[a][b] = 1;
  cout << solve(c,d) << endl;
}