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 long double PI = 3.141592653589793238462643383279502884L;
const long double E = 2.718281828459045235360287471352662498L;
const long double EPS = 1e-12L;
const int64_t mod = 1000000007;
const int s = 100200;
int64_t dp[2][2][2 * s + 1] = {};
int a, b, k, t;
int64_t res = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(11);
  cout.setf(ios::fixed);
  cin >> a >> b >> k >> t;
  fill(dp[0][0] + s + a, dp[0][0] + 2 * s + 1, 1);
  fill(dp[1][0] + s + b, dp[1][0] + 2 * s + 1, 1);
  for (int l = 0; l < 2; ++l) {
    for (int j = 0; j < t; ++j) {
      dp[l][(j + 1) & 1][0] = dp[l][j & 1][k];
      for (int i = -s + 1; i <= s; ++i) {
        dp[l][(j + 1) & 1][s + i] = dp[l][(j + 1) & 1][s + i - 1];
        dp[l][(j + 1) & 1][s + i] +=
            dp[l][j & 1][(i + k <= s ? s + i + k : 2 * s)];
        dp[l][(j + 1) & 1][s + i] -=
            i - k - 1 >= -s ? dp[l][j & 1][s + i - k - 1] : 0;
        dp[l][(j + 1) & 1][s + i] =
            (dp[l][(j + 1) & 1][s + i] % mod + mod) % mod;
      }
    }
  }
  for (int i = -s + 1; i <= s; ++i) {
    res += (dp[0][t & 1][s + i] - dp[0][t & 1][s + i - 1]) *
           dp[1][t & 1][s + i - 1];
    res = (res % mod + mod) % mod;
  }
  cout << res << endl;
  return 0;
}
