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: Wrong Answer)


/*
LANG:C++
PROB:

*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
#include <iostream>
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)


using namespace __gnu_pbds;
using namespace std;


typedef tree<int, null_type, less<int>, rb_tree_tag,
            tree_order_statistics_node_update>
    ordered_set;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
typedef unsigned long long ui64;

template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
const int mod=998244353;
  i64 poww(i64 a, i64 b)
  {
    a%=mod;
    i64 res=1;
    while(b)
      {
if(b&1)res=res*a%mod;

      a=a*a%mod;
      b/=2;

      }
      return res;
  }
  i64 inv(i64 a)
  {
    return poww(a,mod-2);
  }
  i64 dp[3001][3001];
  i64 rec(i64 a,i64 b,i64 c,i64 d)
  {
    if(a==c&&b==d)return 1;
    if(dp[a][b]!=-1)return dp[a][b];

    i64 k=0;
    if(a<c)
    k=rec(a+1,b,c,d)*b,k%=mod;
    if(b<d)
    k+=rec(a,b+1,c,d)*a,k%=mod;
    if(a<c&&b<d)
    k-=rec(a+1,b+1,c,d)*a*b;
    k%=mod;
    if(k<0)k+=mod;

    return dp[a][b]=k;
  }


  int main()
  {
ios::sync_with_stdio(false);
    cin.tie(nullptr);

    // ifstream fin(".in");
    // ofstream fout(".in");
    i64 a,b,c,d;
    cin>>a>>b>>c>>d;
    memset(dp,-1,sizeof(dp));

rec(a,b,c,d);
cout<<dp[a][b];
    return 0;
  }