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.io.PrintWriter;
import java.util.*;

public class D {

	static Scanner sc = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);

	static void exit() {
		sc.close();
		out.close();
	}

	int mod = 1000000007;
	int a, b, k, t;
	int M;
	int c [];
	int p [];
	public D () {
		a = sc.nextInt();
		b = sc.nextInt();
		k = sc.nextInt();
		t = sc.nextInt();
		M = Math.abs(2*t*k + a + b+k);
		c = new int [2*M+1];
		p = new int [2*M+1];
		Arrays.fill(c, 0);
		Arrays.fill(p, 0);
		// solve t = 0: c[0][m] = 1 if 0 <= m
		for (int m = 0; m <= M; m++) c[M+m] = 1;
		p[M+0] = c[M+0];
		for (int m = 1; m <= M; m++) p[M+m] = p[M+m-1] + c[M+m];
		
		// solve for t > 0
		for (int n = 1; n <= 2*t; n++) {
			int tp [] = Arrays.copyOf(p, p.length);
			Arrays.fill(c, 0);
			Arrays.fill(p, 0);
			for (int m = -n*k; m <= n*k; m++) {
				c[M+m] = (tp[M+m+k] - tp[M+m-k-1])%mod; 
				if (c[M+m]<0) c[M+m] += mod;
			}
			for (int m = n*k+1; M+m <= 2*M; m++) c[M+m] = c[M+m-1];
			p[M-n*k] = c[M-n*k];
			for (int m = -n*k+1; m <= M; m++) p[M+m] = (p[M+m-1] + c[M+m])%mod;
		}
		/*
		for (int n = 0; n <= 2; n++) {
			for (int m = -n*k; m <= n*k; m++) {
				out.println("c[" + n + "][" + m + "]=" + c[n][M+m]);
			}
		}
		*/
		out.println(c[M+a-b-1]);
		//out.println(c(2*T, a-b-1));
	}

	/*
	int c(int n, int m) {
		if (n == 0) {
			return 0<=m ? 1 : 0;
		}
		if (m < -n*k) return 0;
		if (m > n*k) return c(n, n*k);
		if (c[n][M+m] != -1) return c[n][M+m];
		c[n][M+m] = p(n-1, m+k) - p(n-1, m-k-1);
		//out.println("c["+n+"]["+m+"]=" + c[n][M+m]);
		return c[n][M+m];
	}
	
	int p(int n, int m) {
		if (m < -n*k) return 0;
		if (p[n][M+m] != -1) return p[n][M+m];
		p[n][M+m] = p(n, m-1) + c(n, m);
		//out.println("p["+n+"]["+m+"]=" + p[n][M+m]);
		return p[n][M+m];
	}
	*/
	
	public static void main(String[] args) {
		new D();
		exit();
	}
}