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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;


public class D {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int a = nextInt();
		int b = nextInt();
		int k = nextInt();
		int t = nextInt();
		int all = 2*k*t;
		int[]dpM1 = new int[all+1];
		int[]dpM2 = new int[all+1];
		int mod = (int) (1e9+7);
		dpM1[0] = 1;
		int[]sum = new int[all+1];
		sum[0] = 1;
		for (int i = 1; i <= all; i++) {
			sum[i] = sum[i-1]+dpM1[i];
		}
		for (int i = 1; i <= t; i++) {
			Arrays.fill(dpM2, 0);
			for (int j = 0; j <= all; j++) {
				dpM2[j] = sum[j];
				if (j-2*k-1 >= 0)
					dpM2[j] -= sum[j-2*k-1];
				if (dpM2[j] < 0)
					dpM2[j] += mod;
			}
			dpM1 = dpM2.clone();
			sum[0] = dpM1[0];
			for (int j = 1; j <= all; j++) {
				sum[j] = sum[j-1] + dpM1[j];
				sum[j] %= mod;
			}
		}
		int[]dpL1 = new int[all+1];
		int[]dpL2 = new int[all+1];
		dpL1[0] = 1;
		sum[0] = 1;
		for (int i = 1; i <= all; i++) {
			sum[i] = sum[i-1]+dpL1[i];
		}
		for (int i = 1; i <= t; i++) {
			Arrays.fill(dpL2, 0);
			for (int j = 0; j <= all; j++) {
				dpL2[j] = sum[j];
				if (j-2*k-1 >= 0)
					dpL2[j] -= sum[j-2*k-1];
				if (dpL2[j] < 0)
					dpL2[j] += mod;
			}
			dpL1 = dpL2.clone();
			sum[0] = dpL1[0];
			for (int j = 1; j <= all; j++) {
				sum[j] = sum[j-1] + dpL1[j];
				sum[j] %= mod;
			}
		}
		long ans = 0;
		long[] sumL = new long[all+1];
		sumL[0] = dpL1[0];
		for (int i = 1; i <= all; i++) {
			sumL[i] = (sumL[i-1]+dpL1[i]) % mod;
		}
		for (int i = 0; i <= all; i++) {
			if (i+a-b-1 >= 0)
				ans += dpM1[i] * sumL[Math.min(all, i+a-b-1)];
			ans %= mod;
		}
		System.out.println(ans);
		pw.close();
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}