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>
#pragma GCC optimize(3)
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
const long long mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template <typename T>
inline void read(T &a) {
  char c = getchar();
  T x = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  a = f * x;
}
int f[N];
int sum[N];
int a, b, k, t;
int main() {
  read(a), read(b), read(k), read(t);
  int p = 2 * k * t;
  f[p] = 1;
  for (int i = 1; i <= t * 2; i++) {
    sum[0] = f[0];
    for (int j = 1; j <= p * 2 + k; j++) sum[j] = (sum[j - 1] + f[j]) % mod;
    for (int j = 0; j <= k; j++) f[j] = sum[j + k];
    for (int j = k + 1; j <= p * 2; j++)
      f[j] = (sum[j + k] - sum[j - k - 1] + mod) % mod;
  }
  int ans = 0;
  for (int i = p - (a - b) + 1; i <= p * 2; i++) ans = (ans + f[i]) % mod;
  printf("%d\n", ans);
  ;
  return 0;
}
