Memory and his friend Lexa are competing to get higher score in one popular computer game. Memory starts with score a and Lexa starts with score b. In a single turn, both Memory and Lexa get some integer in the range [ - k;k] (i.e. one integer among  - k, - k + 1, - k + 2, ..., - 2, - 1, 0, 1, 2, ..., k - 1, k) and add them to their current scores. The game has exactly t turns. Memory and Lexa, however, are not good at this game, so they both always get a random integer at their turn.

Memory wonders how many possible games exist such that he ends with a strictly higher score than Lexa. Two games are considered to be different if in at least one turn at least one player gets different score. There are (2k + 1)2t games in total. Since the answer can be very large, you should print it modulo 109 + 7. Please solve this problem for Memory.

Input

The first and only line of input contains the four integers a, b, k, and t (1 ≤ a, b ≤ 100, 1 ≤ k ≤ 1000, 1 ≤ t ≤ 100) — the amount Memory and Lexa start with, the number k, and the number of turns respectively.

Output

Print the number of possible games satisfying the conditions modulo 1 000 000 007 (109 + 7) in one line.

Examples

Input

1 2 2 1


Output

6


Input

1 1 1 2


Output

31


Input

2 12 3 1


Output

0

Note

In the first sample test, Memory starts with 1 and Lexa starts with 2. If Lexa picks  - 2, Memory can pick 0, 1, or 2 to win. If Lexa picks  - 1, Memory can pick 1 or 2 to win. If Lexa picks 0, Memory can pick 2 to win. If Lexa picks 1 or 2, Memory cannot win. Thus, there are 3 + 2 + 1 = 6 possible games in which Memory wins.
import java.util.Scanner;

public class main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		final int mod = 1000000007;
		int a,b,k,t;
		Scanner in = new Scanner(System.in);
		a = in.nextInt();
		b = in.nextInt();
		k = in.nextInt();
		t = in.nextInt();
		in.close();
		long[] pred = new long[4*k*(t+1)+1];
		long[] tek  = new long[4*k*(t+1)+1];
		long[] temp;
		int offset = 2*k*(t+1);
		int diff = a - b;
		tek[offset] = 1;
		
		for(int i = 0; i < 2*t; i++)
		{
			temp = pred;
			pred = tek;
			tek = temp;
			long tekelem = 0;
			for(int j = offset-k*(i+1); j<=offset;j++)
			{
				tekelem = (tekelem+pred[j+k])%mod;
				tekelem = (tekelem-pred[j-k-1])%mod;
				tek[j] = tekelem;
			}
			for(int j = offset+1; j<=offset+k*(i+1);j++)
			{
				tek[j] = tek[2*offset-j];
			}
			
			//System.out.println(i/2);
		}
		long globsum = 0;
		for(int i = offset-diff+1; i < 4*k*(t+1)+1; i++)
		{
			if ((i>=0) && (i<4*k*(t+1)+1))
			{
				globsum = (globsum+tek[i])%mod;
			}
		}
		System.out.println(globsum);
	}

}