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;
const int md = 1e9 + 7;
const int maxn = 2e5 + 100;
int n, kk;
long long f[maxn * 2 + 5][2];
int main() {
  int T, a, b, poi = 0;
  long long now;
  scanf("%d%d%d%d", &a, &b, &kk, &T);
  kk *= 2;
  n = T * kk + 100;
  f[a - b + n + 1][0] = 1;
  for (int i = a - b + n + 2; i <= 2 * n + 1; ++i) f[i][0] = 1;
  for (int t = 1; t <= T; ++t) {
    now = 0;
    for (int i = kk + 1; i >= 1; --i) now += f[i][poi];
    now %= md;
    f[1][poi ^ 1] = now;
    for (int i = 2; i <= 2 * n + 1; ++i) {
      int l = max(1, i - kk - 1);
      now = (now - f[i - 1][poi] + f[l - 1][poi] + md);
      l = min(2 * n + 1, i + kk);
      now = (now + f[l][poi] - f[i - 1][poi] + md);
      now %= md;
      f[i][poi ^ 1] = (f[i - 1][poi ^ 1] + now) % md;
    }
    poi ^= 1;
  }
  long long ans = (f[2 * n + 1][poi] - f[n + 1][poi] + md) % md;
  printf("%lld\n", ans);
  return 0;
}
