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.InputStreamReader;
import java.util.Arrays;

public class E {

	public static void main(String[] args) throws Exception {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		String l[] = bf.readLine().split(" ");
		int max, min;
		int a = Integer.parseInt(l[0]);
		int b = Integer.parseInt(l[1]);
		int K = Integer.parseInt(l[2]);
		int T = Integer.parseInt(l[3]);
		int S = K * T;
		int prevSums[] = new int[S * 2 + 1];
		int DP[][] = new int[T + 1][S * 2 + 1];
		int mod = 1000000000 + 7;
		DP[0][S] = 1;
		for (int t = 1; t <= T; t++) {
			prevSums[0] = DP[t - 1][0];
			for (int s = 1; s < prevSums.length; s++) {
				prevSums[s] = (prevSums[s - 1] + DP[t - 1][s]) % mod;
				prevSums[s] %= mod;
			}
			for (int s = -S; s <= S; s++) {
				max = Math.min(K, s + S);
				min = Math.max(-K, s - S);
				DP[t][s + S] += prevSums[s - min + S];
				DP[t][s + S] %= mod;

				if (s - max + S > 0) {
					DP[t][s + S] -= prevSums[s - max + S - 1];
					while (DP[t][s + S] < 0)
						DP[t][s + S] += mod;
					DP[t][s + S] %= mod;
				}
			}
			Arrays.fill(prevSums, 0);
		}
		for (int i = -S; i <= S; i++) {
			prevSums[i + S] = DP[T][i + S];
			if (i + S > 0)
				prevSums[i + S] += prevSums[i + S - 1];
			prevSums[i + S] %= mod;
		}
		int lo, hi, mid;
		long ans = 0;
		long tmp1, tmp2;
		for (int i = -S; i <= S; i++) {
			lo = 0;
			hi = prevSums.length;
			while (lo + 1 < hi) {
				mid = (lo + hi) >> 1;
				if (mid - S + b < i + a) {
					lo = mid;
				} else {
					hi = mid;
				}
			}
			int j = lo - S;
			if (i + a > j + b) {
				tmp1 = DP[T][i + S];
				tmp2 = prevSums[j + S];
				ans += tmp1 * tmp2;
				ans %= mod;
			}
		}
		System.out.println(ans);
	}
}