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;
inline int Min(int &a, int b) { return a > b ? a = b : a; }
inline int Max(int &a, int b) { return a < b ? a = b : a; }
inline int min(int a, int b) { return a > b ? b : a; }
inline int max(int a, int b) { return a < b ? b : a; }
inline int abs(int a) { return a > 0 ? a : -a; }
const int INF = 1 << 29;
const int mod = 1e9 + 7, N = 500000;
int fac[N], rev[N];
int Pow(int a, int b) {
  int res = 1;
  for (b; b; b >>= 1, a = a * 1ll * a % mod)
    if (b & 1) res = res * 1ll * a % mod;
  return res;
}
int C(int n, int m) {
  if (m > n) return 0;
  int res;
  res = 1ll * fac[n] * rev[m] % mod * rev[n - m] % mod;
  return res;
}
int a, b, k, t;
int main() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  rev[N - 1] = Pow(fac[N - 1], mod - 2);
  rev[0] = 1;
  for (int i = N - 2; i; i--) rev[i] = rev[i + 1] * (i + 1ll) % mod;
  scanf("%d%d%d%d", &a, &b, &k, &t);
  int ans = 0;
  for (int j = 0; j <= 2 * t; j++)
    ans = (ans +
           1ll * Pow(-1, j) * C(2 * t, j) % mod *
               (1ll * C(210000 + 2 * k * t - a + b + 2 * t - 1 -
                            (2 * k + 1) * j + 1,
                        2 * t) -
                C(1 + 2 * k * t - a + b + 2 * t - 1 - (2 * k + 1) * j, 2 * t)) %
               mod) %
          mod;
  ans = (ans % mod + mod) % mod;
  printf("%d\n", ans);
  return 0;
}
