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 N = 434567;
const int B = 211001;
const int T = 123;
const int MOD = 1e9 + 7;
int pref[N], f[T][N], a, b, k, t;
int main() {
  ios::sync_with_stdio(false);
  cin >> a >> b >> k >> t;
  f[0][a - b + B] = 1;
  for (int i = 0; i < 2 * k + 1; i++) {
    pref[i + a - b + B]++;
  }
  for (int i = 1; i <= t; i++) {
    int pyramid = 0;
    for (int j = -2 * k; j <= 2 * k; j++) {
      pyramid = (((long long)(2 * k - abs(j) + 1) *
                  f[i - 1][-2 * k * i + a - b + j + B] % MOD) +
                 pyramid) %
                MOD;
    }
    for (int j = -2 * k * i + a - b; j <= 2 * k * i + a - b; j++) {
      f[i][j + B] = pyramid;
      pyramid = (pyramid - pref[j + B] + MOD) % MOD;
      pyramid = (pyramid + pref[j + 2 * k + 1 + B]) % MOD;
    }
    for (int j = -2 * k * (i + 1) + a - b; j <= 2 * k * (i + 1) + a - b; j++) {
      pref[j + B] = (pref[j - 1 + B] + f[i][j + B]) % MOD;
    }
    for (int j = 2 * k * (i + 1) + a - b; j >= -2 * k * (i + 1) + a - b; j--) {
      pref[j + B] = (pref[j + B] - pref[j - 2 * k - 1 + B] + MOD) % MOD;
    }
  }
  int sum = 0;
  for (int i = B + 1; i < N; i++) {
    sum = (sum + f[t][i]) % MOD;
  }
  printf("%d\n", sum);
  return 0;
}
