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 C[2][401005];
int qzh[2][401005];
int a, b, k, t;
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  int Z = 2 * t * k + 3;
  C[0][Z] = 1;
  int cur = 1;
  for (int i = 1; i <= 2 * t; i++) {
    int pre = (cur + 1) % 2;
    memset(C[cur], 0, sizeof(C[cur]));
    memset(qzh[pre], 0, sizeof(pre));
    int tt = i * k;
    for (int j = -2 * k * t; j <= 2 * k * t; j++)
      qzh[pre][j + Z] = (qzh[pre][j - 1 + Z] + C[pre][j + Z]) % 1000000007ll;
    for (int j = -tt; j <= tt; j++) {
      int r = min(j + k, tt);
      int l = max(j - k, -tt);
      int p = (qzh[pre][r + Z] - qzh[pre][j - k - 1 + Z] + 1000000007ll) %
              1000000007ll;
      C[cur][j + Z] += p;
      C[cur][j + Z] %= 1000000007ll;
    }
    if (i == 2 * t) {
      memset(qzh[cur], 0, sizeof(qzh[cur]));
      for (int j = -tt; j <= tt; j++)
        qzh[cur][j + Z] = (qzh[cur][j - 1 + Z] + C[cur][j + Z]) % 1000000007ll;
    }
    cur = pre;
  }
  int yy;
  int p = a - b;
  yy = -p + Z;
  cur = (cur + 1) % 2;
  int ans = 0;
  if (yy <= 2 * k * t + Z)
    ans =
        (qzh[cur][2 * k * t + Z] - qzh[cur][yy] + 1000000007ll) % 1000000007ll;
  printf("%d\n", ans);
  return 0;
}
