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>
static const int MAXK = 1003;
static const int MAXT = 103;
static const int MODULUS = 1e9 + 7;
static const int MAXPWR = 4 * MAXK * MAXT;
inline int fpow(long long base, int exp) {
  long long ans = 1;
  for (; exp; exp >>= 1) {
    if (exp & 1) ans = ans * base % MODULUS;
    base = base * base % MODULUS;
  }
  return (int)ans;
}
inline int inv(int x) { return fpow(x, MODULUS - 2); }
int fact[MAXPWR], fact_inv[MAXPWR];
inline int nCr(int n, int m) {
  if (n < 0 || m < 0 || n - m < 0) return 0;
  return (long long)fact[n] * fact_inv[m] % MODULUS * fact_inv[n - m] % MODULUS;
}
int a, b, k, t;
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  fact[0] = fact_inv[0] = fact[1] = fact_inv[1] = 1;
  int p = 2 * t, n = 2 * t;
  for (int i = 2; i < MAXPWR; ++i) {
    fact[i] = (long long)fact[i - 1] * i % MODULUS;
    fact_inv[i] = inv(fact[i]);
  }
  long long ans = 0;
  for (int i = 0; i <= 2 * t; ++i) {
    ans = (ans + (long long)nCr(p, i) *
                     (nCr(p + 4 * k * t - (2 * k + 1) * i, p) -
                      nCr(p + 2 * k * t + b - a - (2 * k + 1) * i, p)) %
                     MODULUS * (i % 2 == 0 ? 1 : -1)) %
          MODULUS;
  }
  printf("%I64d\n", (ans % MODULUS + MODULUS) % MODULUS);
  return 0;
}
