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 = 1000000007;
const int size = 200000 + 4000;
int arrMemory[2][size];
int arrLexa[2][size];
int positive_modulo(int i, int n) { return (i % n + n) % n; }
int main() {
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  arrMemory[0][a + 100001] = 1;
  arrLexa[0][b + 100001] = 1;
  for (int i = 0; i < t; i++) {
    int accum = 0;
    int accum2 = 0;
    for (int j = 0; j < size; j++) {
      accum = (accum + arrMemory[0][j]) % MOD;
      arrMemory[1][j] = accum;
      accum2 = (accum2 + arrLexa[0][j]) % MOD;
      arrLexa[1][j] = accum2;
    }
    for (int j = 0; j < size; j++) {
      arrMemory[0][j] = positive_modulo(
          arrMemory[1][min(j + k, size - 1)] - arrMemory[1][max(j - k - 1, 0)],
          MOD);
      arrLexa[0][j] = positive_modulo(
          arrLexa[1][min(j + k, size - 1)] - arrLexa[1][max(j - k - 1, 0)],
          MOD);
    }
  }
  long long cnt = 0;
  int accum2 = 0;
  for (int j = 0; j < size; j++) {
    accum2 = (accum2 + arrLexa[0][j]) % MOD;
    arrLexa[1][j] = accum2;
  }
  for (int j = 1; j < size; j++) {
    long long temp = (arrMemory[0][j] % MOD);
    temp *= arrLexa[1][j - 1];
    cnt = (cnt + (temp % MOD)) % MOD;
  }
  cout << cnt;
}
