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 N = 3010;
const int P = 998244353;

int mult(int a, int b) {
    return a * 1ll * b % P;
}
void add(int &a, int b) {
    a += b;
    if(a >= P) a -= P;
}
void sub(int &a, int b) {
    a -= b;
    if(a < 0) a += P;
}

int dp[N][N];

int main() {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    dp[a][b] = 1;
    for (int i = a; i <= c; ++i) {
        for (int j = b; j <= d; ++j) {
            add(dp[i][j], mult(j, dp[i-1][j]));
            add(dp[i][j], mult(i, dp[i][j-1]));
            sub(dp[i][j], mult(mult(i - 1, j - 1), dp[i-1][j-1]));

        }
    }
    printf("%d\n", dp[c][d]);
}