Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

1 1 2 2

Sample Output 1

3

Any two of the three squares other than the bottom-left square can be painted black.

Sample Input 2

2 1 3 4

Sample Output 2

65

Sample Input 3

31 41 59 265

Sample Output 3

387222020

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
typedef long long ll;
typedef long double ld;
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define mod 998244353
#define inf 1000000000000000000
#define bpc(x) __builtin_popcountll(x)
#define autoit(x,it) for(auto it = x.begin(); it != x.end(); it++)
#define rep(n) for(ll i = 0; i < n; i++)
#define repi(i,n) for(ll i = 0; i < n; i++)

#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 

#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 

using namespace std;

#define N 3005

ll dp[N][N];
ll a,b;
ll get(ll i, ll j)
{
    if(i<a || j<b)
        return 0;
    if(dp[i][j]!=-1)
        return dp[i][j];
    if(i == a && j == b)
        return dp[i][j] = 1;
    return dp[i][j] = ((get(i-1,j)*(j))%mod + (get(i,j-1)*(i))%mod - (get(i-1,j-1)*(i-1)*(j-1))%mod + mod)%mod;    
}

int main()
{
  FAST/**/
   
   memset(dp,-1,sizeof(dp));
   ll c,d;
   cin>>a>>b>>c>>d;
   
   ll ans = get(c,d);
   cout<<ans;
   return 0;        
}