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 signed int MAXT = 100 + 5;
const signed int MAXS = (1000 << 1) * MAXT + 100 + 5;
const signed int MOD = 1e9 + 7;
signed int dp1[2][MAXS];
signed int dp2[2][MAXS];
bool idx;
void possible(signed int s, signed int k, signed int t,
              signed int dp[2][MAXS]) {
  dp[0][s] = 1;
  idx = 0;
  signed int l = k << 1;
  for (signed int i = (1); i <= (t); ++i) {
    signed int limit = (k << 1) * i + s;
    signed int prev = 0;
    for (signed int j = (0); j <= (l); ++j) {
      prev += dp[idx][j];
      if (prev >= MOD) prev -= MOD;
      dp[!idx][j] = prev;
    }
    for (signed int j = (l + 1); j <= (limit); ++j) {
      prev += dp[idx][j] - dp[idx][j - l - 1];
      if (prev >= MOD) prev -= MOD;
      if (prev < 0) prev += MOD;
      dp[!idx][j] = prev;
    }
    idx = !idx;
  }
}
signed int main() {
  signed int a, b, k, t;
  cin >> a >> b >> k >> t;
  possible(a, k, t, dp1);
  bool idx1 = idx;
  possible(b, k, t, dp2);
  bool idx2 = idx;
  signed int l1 = (k << 1) * t + a;
  signed int l2 = (k << 1) * t + b;
  signed int ans = 0;
  signed int prev = 0;
  signed int s = max(a, b + 1);
  signed int j = b;
  for (signed int i = (s); i <= (l1); ++i) {
    while (j < i && j <= l2) {
      prev += dp2[idx2][j++];
      if (prev >= MOD) prev -= MOD;
    }
    ans = (ans + 1LL * dp1[idx1][i] * prev) % MOD;
  }
  cout << ans << endl;
  return 0;
}
