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 powr(long long a, long long b, long long mod) {
  long long x = 1 % mod;
  a %= mod;
  while (b) {
    if (b & 1) x = (x * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return x;
}
long long inv(long long a, long long mod) { return powr(a, mod - 2, mod); }
const int maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
const int N = 500005;
long long fact[N], invfact[N];
long long C(int n, int k) {
  if (n < 0 || k < 0 || n < k) return 0;
  return (fact[n] * ((invfact[k] * invfact[n - k]) % mod)) % mod;
}
int main() {
  int a, b, k, t;
  scanf("%d", &(a)), scanf("%d", &(b)), scanf("%d", &(k)), scanf("%d", &(t));
  fact[0] = invfact[0] = 1;
  for (int i = 1; i < N; i++)
    fact[i] = (fact[i - 1] * i) % mod, invfact[i] = inv(fact[i], mod);
  long long ans = 0;
  for (int r = 0; r <= 2 * t; r++)
    for (int i = 0; i <= 4 * k * t; i++)
      if ((r * (2 * k + 1) + i > b - a + 2 * k * t) &&
          (r * (2 * k + 1) + i <= 4 * k * t)) {
        if (!(r & 1))
          ans = (ans + C(2 * t, r) * C(2 * t + i - 1, i)) % mod;
        else
          ans = (ans - C(2 * t, r) * C(2 * t + i - 1, i)) % mod;
      }
  ans = (ans % mod + mod) % mod;
  cout << ans;
}
