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)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define rept(i,k,n) for(ll i = (k); i < (ll)(n); i++)
#define all(v) v.begin(), v.end()
#define Sort(v) sort(v.begin(),v.end())
#define Reverse(v) reverse(v.begin(),v.end())
typedef long long ll;
typedef pair<int,int> Pii;
typedef pair<long long,long long> Pll;
typedef tuple<int,int,int> Tii;

template <typename T>
ostream& operator<<(ostream& s, const vector<T> &vec){
    for(int i=0;i<vec.size();i++){
        if(i) s << ',';
        else s << '{';
        s <<setw(2)<< vec[i];
    }
    s << '}' << endl;
    return s;
}
ll mod = 998244353;

int main(){
    ll a,b,c,d;
    cin>>a>>b>>c>>d;
    vector<vector<ll>> dp(c+1,vector<ll>(d+1,0));
    dp[a][b]=1;
    rept(i,a,c+1){
        rept(j,b,d+1){
            if(i==a&&j==b) continue;
            dp[i][j]=(dp[i-1][j]+dp[i][j-1]+max(dp[i-1][j]*(j-1),dp[i][j-1]*(i-1)))%mod;
        
        }
    }
  //  cout<<dp[c][d]<<endl;
    cout<<dp<<endl;
    return 0;
}