#arc133e. [arc133_e]Cyclic Medians

[arc133_e]Cyclic Medians

Problem Statement

Given are integers NN, MM, VV, and AA. Consider the following procedure.

  • Choose a sequence of NN integers between 11 and VV (inclusive): x=(x1,x2,cdots,xN)x=(x_1,x_2,\\cdots,x_N).
  • Choose a sequence of MM integers between 11 and VV (inclusive): y=(y1,y2,cdots,yM)y=(y_1,y_2,\\cdots,y_M).
  • Let aa be a variable and initialize it with a=Aa=A.
  • For each i=0,1,cdots,NtimesM1i=0,1,\\cdots,N \\times M-1, do the following.
    • Replace the value of aa with the median of a,x(ibmodN)+1,y(ibmodM)+1a,x_{(i \\bmod N)+1},y_{(i \\bmod M)+1}.
  • Print the final value of aa.

Consider doing this procedure with every possible pair of sequences x,yx,y. Find the sum of the values that will be printed, modulo 998244353998244353.

Constraints

  • 1leqN,Mleq2000001 \\leq N,M \\leq 200000
  • 1leqAleqVleq2000001 \\leq A \\leq V \\leq 200000
  • All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN MM VV AA

Output

Print the answer.


Sample Input 1

2 1 2 1

Sample Output 1

11

For example, when x=(1,2),y=(2)x=(1,2),y=(2), the procedure goes as follows.

  • Initialize aa with a=1a=1.
  • For i=1i=1: replace the value of aa with the median of a(=1),x1(=1),y1(=2)a(=1),x_1(=1),y_1(=2), which is 11.
  • For i=2i=2: replace the value of aa with the median of a(=1),x2(=2),y1(=2)a(=1),x_2(=2),y_1(=2), which is 22.
  • Print a(=2)a(=2).

There are three cases where 22 will be printed: (x=(1,2),y=(2))(x=(1,2),y=(2)), (x=(2,1),y=(2))(x=(2,1),y=(2)), (x=(2,2),y=(2))(x=(2,2),y=(2)). In the other five cases, 11 will be printed. Therefore, the answer is 2times3+1times5=112 \\times 3 + 1\\times 5=11.


Sample Input 2

2 2 5 4

Sample Output 2

2019

Sample Input 3

2100 2300 2201 2022

Sample Output 3

407723438