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>
int p1, p2, f[500210], i, j, sum[2][500210], a, b, k, t, mo = 1e9 + 7, ans;
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  f[a - b + 200100] = 1;
  t <<= 1;
  for (i = a - b + 200100; i <= k * t + a - b + 200100; i++) sum[0][i] = 1;
  p2 = 1;
  for (i = 1; i <= t; i++) {
    std::swap(p1, p2);
    memset(f, 0, sizeof(f));
    memset(sum[p1], 0, sizeof(sum[p1]));
    for (j = 0; j <= k * t + a - b + 200100; j++) {
      f[j] = sum[p2][std::min(j + k, k * t + a - b + 200100)] -
             sum[p2][std::max(j - k - 1, 0)];
      if (f[j] >= mo)
        f[j] -= mo;
      else if (f[j] < 0)
        f[j] += mo;
      sum[p1][j] = sum[p1][std::max(j - 1, 0)] + f[j];
      if (sum[p1][j] >= mo)
        sum[p1][j] -= mo;
      else if (sum[p1][j] < 0)
        sum[p1][j] += mo;
    }
  }
  for (i = 200101; i <= k * t + a - b + 200100; i++) {
    ans += f[i];
    if (ans >= mo) ans -= mo;
  }
  printf("%d", ans);
  return 0;
}
