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.
#include <bits/stdc++.h>
using namespace std;
int A, B, K, T;
int Sum[2][2 * 4 * 1000 * (100 + 5) + 5];
int main() {
  scanf("%d%d%d%d", &A, &B, &K, &T);
  for (int j = B - A; j <= 2 * K * T; j++) Sum[0][j + 4 * 1000 * (100 + 5)] = 1;
  for (int i = 1; i <= 2 * T; i++)
    for (int j = -4 * K * T; j <= 4 * K * T; j++)
      Sum[i % 2][j + 4 * 1000 * (100 + 5)] =
          (Sum[i % 2][j - 1 + 4 * 1000 * (100 + 5)] +
           (Sum[(i - 1) % 2][j + K + 4 * 1000 * (100 + 5)] -
            Sum[(i - 1) % 2][j - K - 1 + 4 * 1000 * (100 + 5)]) %
               1000000007) %
          1000000007;
  int Ans = 0;
  for (int i = -4 * K * T; i <= -1; i++)
    Ans = (Ans + (Sum[(2 * T - 1) % 2][i + K + 4 * 1000 * (100 + 5)] -
                  Sum[(2 * T - 1) % 2][i - K - 1 + 4 * 1000 * (100 + 5)]) %
                     1000000007) %
          1000000007;
  printf("%d", (Ans + 1000000007) % 1000000007);
}
