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 llint = long long;
using namespace std;
const int K = 1005;
const int T = 101;
const int N = 4 * K * T + 5 * K;
const int MOD = int(1e9) + 7;
int madd(int a, int b) { return (a + b) % MOD; }
int msub(int a, int b) { return ((llint)a - b + MOD) % MOD; }
int mmul(int a, int b) { return (llint)a * b % MOD; }
int dp[N];
int cum[N + 1];
int &Cum(int i) { return cum[N / 2 + i]; }
int Sum(int a, int b) { return msub(Cum(b), Cum(a)); }
int &Dp(int i) { return dp[N / 2 + i]; }
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  int diff = a - b;
  int min = diff - 2 * k * t;
  int max = diff + 2 * k * t;
  for (int i = (0); i < (N); ++i) {
    dp[i] = 0;
  }
  Dp(diff) = 1;
  cum[0] = 0;
  for (int i = (1); i < (N + 1); ++i) {
    cum[i] = madd(cum[i - 1], dp[i - 1]);
  }
  for (int i = (0); i < (t); ++i) {
    int wind = 0;
    for (int kk = (-2 * k); kk < (2 * k + 1); ++kk) {
      wind = madd(wind, mmul(Dp(min + kk), (2 * k - abs(kk) + 1)));
    }
    for (int j = (min); j < (max + 1); ++j) {
      Dp(j) = wind;
      wind = madd(msub(wind, Sum(j - 2 * k, j + 1)), Sum(j + 1, j + 2 * k + 2));
    }
    cum[0] = 0;
    for (int i = (1); i < (N + 1); ++i) {
      cum[i] = madd(cum[i - 1], dp[i - 1]);
    }
  }
  cout << Sum(1, max + 1) << endl;
  return 0;
}
