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 mod = 1e9 + 7;
const int N = 2e5 + 200;
long long dpa[2][N];
long long dpb[2][N];
int main() {
  int a, b, k, t, mn, mx, l, r;
  cin >> a >> b >> k >> t;
  mn = min(a, b) - k * t;
  mx = max(a, b) + k * t;
  dpa[0][a - mn] = 1;
  for (int i = 1; i <= t; i++) {
    for (int j = 0; j <= mx - mn; j++) dpa[1][j] = 0;
    for (int j = 0; j <= mx - mn; j++) {
      if (dpa[0][j] == 0) continue;
      l = max(0, j - k);
      r = min(mx - mn + 1, j + k + 1);
      dpa[1][l] += dpa[0][j];
      dpa[1][r] -= dpa[0][j];
    }
    for (int j = 0; j <= mx - mn; j++) {
      if (j > 0) dpa[1][j] = (dpa[1][j] % mod + dpa[1][j - 1] % mod) % mod;
      dpa[0][j] = dpa[1][j];
    }
  }
  dpb[0][b - mn] = 1;
  for (int i = 1; i <= t; i++) {
    for (int j = 0; j <= mx - mn; j++) dpb[1][j] = 0;
    for (int j = 0; j <= mx - mn; j++) {
      if (dpb[0][j] == 0) continue;
      l = max(0, j - k);
      r = min(mx - mn + 1, j + k + 1);
      dpb[1][l] += dpb[0][j];
      dpb[1][r] -= dpb[0][j];
    }
    for (int j = 0; j <= mx - mn + 1; j++) {
      if (j > 0) dpb[1][j] = (dpb[1][j] % mod + dpb[1][j - 1] % mod) % mod;
      dpb[0][j] = dpb[1][j];
    }
  }
  for (int j = 0; j <= mx - mn; j++) {
    if (j > 0) dpb[0][j] = (dpb[0][j] % mod + dpb[0][j - 1] % mod) % mod;
  }
  long long ans = 0;
  for (int i = mn + 1; i <= mx; i++) {
    if (dpa[0][i - mn] == 0) continue;
    long long ctra = dpa[0][i - mn];
    long long ctrb = dpb[0][i - mn - 1];
    long long temp = (ctra % mod * ctrb % mod) % mod;
    ans = (ans % mod + temp % mod) % mod;
  }
  cout << ans << "\n";
  return 0;
}
