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)

/*input

*/


// assic value of ('0'-'9') is(48 - 57) and (a-z) is (97-122) and (A-Z) is(65-90) and 32 for space
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;

// order_of_key (k) : Number of items strictly smaller than k .
// find_by_order(k) : K-th element in a set (counting from zero).
using namespace __gnu_pbds;
template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define ll          long long int
#define pb          push_back
#define pii         pair<ll ,ll >
#define vpii        vector< pii >
#define vi          vector<ll >
#define vs			vector< string >
#define vvi         vector< vector< ll > >
#define inf			(ll)1e18
#define all(it,a)   for(auto it=(a).begin();it!=(a).end();it++) 
#define F           first
#define S           second
#define sz(x)       (ll )x.size()
#define rep(i,a,b)	for(ll  i=a;i<b;i++)
#define repr(i,a,b) for(ll  i=a;i>b;i--)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define mp          make_pair
#define whatis(x)   cout << #x << " is " << x << "\n";
#define graph(n)    adj(n,vector< ll > () )
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

#define debug(x)     cout << #x << " " << x << endl;
#define debug_p(x)   cout << #x << " " << x.F << " " << x.S << endl;
#define debug_v(x)   {cout << #x << " "; for (auto ioi : x) cout << ioi << " "; cout << endl;}
#define debug_vp(x)  {cout << #x << " "; for (auto ioi : x) cout << '[' << ioi.F << " " << ioi.S << ']'; cout << endl;}
#define debug_v_v(x) {cout << #x << "/*\n"; for (auto ioi : x) { for (auto ioi2 : ioi) cout << ioi2 << " "; cout << '\n';} cout << "*/" << #x << endl;}
const ll mod = 998244353;
const ll N = 3005;

vvi dp(N,vi (N,-1));
ll a,b,c,d;

ll fun(ll x,ll y)
{
	if(dp[x][y]!=-1) return dp[x][y];
	if(x==a) dp[x][y] = (fun(x,y-1)*x)%mod;
	else if(y==b) dp[x][y]=(fun(x-1,y)*y)%mod;
	else dp[x][y] = (((fun(x,y-1)*x)%mod + (fun(x-1,y)*y)%mod)%mod - (fun(x-1,y-1)*(x-1)*(y-1))%mod +mod )%mod;
	return dp[x][y];
}
		
int solve()
{
	cin>>a>>b>>c>>d;
	rep(i,0,a+1) rep(j,0,b+1) dp[i][j]=1;
	cout<<fun(c,d);
	return 0;
}

int main()
{
	auto start = chrono::high_resolution_clock::now();

	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	ll test_cases=1;
	//cin>>test_cases;
	while(test_cases--)
		solve();

	auto stop = chrono::high_resolution_clock::now();
	auto duration = chrono::duration_cast<chrono::milliseconds>(stop-start);
	//cout<<"\nduration: "<<(double)duration.count()<<" milliseconds";
}