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;
long long int d[300007][3];
int main() {
  long long int i, j, n, k, a, b, t, ans = 0, p = 1e9 + 7;
  cin >> a >> b >> k >> t;
  d[k * t][0] = 1;
  for (i = 0; i < t; i++) {
    long long int x = 0;
    for (j = 0; j <= k; j++) x = (x + d[j][0]) % p;
    for (j = 0; j <= 2 * k * t; j++) {
      d[j][1] = x;
      x = (x + d[j + k + 1][0]) % p;
      if (j - k >= 0) x = (x - d[j - k][0]) % p;
    }
    for (j = 0; j <= 2 * k * t; j++) d[j][0] = d[j][1];
  }
  long long int dif = abs(a - b);
  for (i = 1; i <= k * t * 2; i++) d[i][0] = (d[i][0] + d[i - 1][0]) % p;
  if (a <= b)
    for (i = dif + 1; i <= 2 * k * t; i++) {
      if (i - dif - 2 >= 0)
        ans = (ans + ((d[2 * k * t][0] - d[i - 1][0]) *
                      (d[i - dif - 1][0] - d[i - dif - 2][0])) %
                         p) %
              p;
      else
        ans = (ans +
               ((d[2 * k * t][0] - d[i - 1][0]) * (d[i - dif - 1][0])) % p) %
              p;
      if (ans < 0) ans += p;
    }
  else {
    for (i = 0; i <= 2 * k * t; i++) {
      if (i == 0)
        ans = (ans + ((d[i][0]) * (d[2 * k * t][0])) % p) % p;
      else
        ans =
            (ans +
             ((d[i][0] - d[i - 1][0]) * (d[2 * k * t][0] - d[i - 1][0])) % p) %
            p;
      if (ans < 0) ans += p;
    }
    dif = a - b - 1;
    for (i = 0; i <= 2 * k * t; i++) {
      if (i > 0)
        ans = (ans + ((d[i][0] - d[i - 1][0]) *
                      (d[min(2 * k * t, i + dif)][0] - d[i][0])) %
                         p) %
              p;
      else
        ans = (ans +
               ((d[i][0]) * (d[min(2 * k * t, i + dif)][0] - d[i][0])) % p) %
              p;
      if (ans < 0) ans += p;
    }
  }
  if (ans < 0) ans += p;
  cout << ans << endl;
}
