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>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
using ll = long long;
using P = pair<int,int>;
using Graph= vector<vector<int>>; 
#define rep(i,n) for (ll i=0; i < (n); ++i)
#define rep2(i,n,m) for(ll i=n;i<=m;i++)
#define rep3(i,n,m) for(ll i=n;i>=m;i--)
#define pb push_back
#define eb emplace_back
#define ppb pop_back
#define fi first 
#define se second
const ll INF=1e18 ;
inline void chmax(ll& a,ll b){a=max(a,b);}
inline void chmin(ll& a,ll b){a=min(a,b);}
// ミント
const int mod = //1e9+7 ;
998244353;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
//昆布
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
}   ;
 
mint modpow(ll a,ll b){
 if(b==0) return 1 ;
 mint c= modpow(a,b/2) ;
 if(b%2==1) return c*c*a ;
 else return c*c ;
}

mint komb(ll n,ll m){
  mint x=1 ;mint y=1 ;
  rep(i,m){
    x*= n-i ;
    y*= i+1 ;
  }
  return x/y ;
}

  mint dp[3005][3005] ;
int main(){
 ll a,b,c,d ;cin>>a>>b>>c>>d ;
 dp[a][b]=1 ;
 rep2(i,a+1,c){
   dp[i][b]= dp[i-1][b]*b ;
 }
 rep2(i,b+1,d){
   dp[a][i]= dp[a][i-1]*a ;
 }
 rep2(i,a+1,c){
   rep2(j,b+1,d){
     dp[i][j]= dp[i-1][j]*j+ dp[i][j-1]*i
     -dp[i-1][j-1]*(i-1)*(j-1) ;
   }
 }
    cout<<dp[c][d]<<endl;
    return 0;
  }