Problem: 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.

Examples

Input

1 1 2 2


Output

3


Input

2 1 3 4


Output

65


Input

31 41 59 265


Output

387222020

Solution (in CPP):

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

const int MOD = 998244353;

int a, b, c, d;
long long dp[3003][3003];

long long solve(int x, int y)
{
    if (x < a || y < b)
        return 0;
    
    if (x == a && y == b)
        return 1;
    
    if (dp[x][y] != -1)
        return dp[x][y];
    
    long long ans = solve(x - 1, y) * y + solve(x, y - 1) * x;
    ans += MOD - solve(x - 1, y - 1) * (x - 1) % MOD * (y - 1) % MOD;

    return dp[x][y] = ans % MOD;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(NULL), cout.tie(NULL);

    cin >> a >> b >> c >> d;
    memset(dp, -1, sizeof(dp));

    cout << solve(c, d) << "\n";
    return 0;
}