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.*;
import java.io.*;

public class Memory_and_Scores {

	public static void main(String[] args) throws IOException {

		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(f.readLine());

		int a = Integer.parseInt(st.nextToken());
		int b = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		int t = Integer.parseInt(st.nextToken());

		f.close();

		long[][] dp = new long[2][k * t * 4 + 401];
		dp[0][k * t * 2 + 200] = 1;

		long mod = 1000000007;

		for (int i = 0; i < 2 * t; i++) {

			int cur = i % 2;
			int nxt = (i + 1) % 2;

			Arrays.fill(dp[nxt], 0);

			for (int j = 0; j <= k * t * 4 + 400; j++) {

				if (dp[cur][j] == 0) {
					continue;
				}

				dp[nxt][j + k + 1] -= dp[cur][j];
				if (dp[nxt][j + k + 1] < 0) {
					dp[nxt][j + k + 1] += mod;
				}
				if (dp[nxt][j + k + 1] >= mod) {
					dp[nxt][j + k + 1] -= mod;
				}

				dp[nxt][j - k] += dp[cur][j];
				if (dp[nxt][j - k] >= mod) {
					dp[nxt][j - k] -= mod;
				}

			}

			for (int j = 1; j <= k * t * 4 + 400; j++) {
				dp[nxt][j] += dp[nxt][j - 1];
				if (dp[nxt][j] >= mod) {
					dp[nxt][j] -= mod;
				}
			}

		}

		int ans = 0;
		for (int i = b - a + k * t * 2 + 201; i <= k * t * 4 + 400; i++) {
			ans += dp[0][i];
			if (ans >= mod) {
				ans -= mod;
			}
		}

		System.out.println(ans);

	}

}
