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;
int a, b, k, t;
int main() {
  cin >> a >> b >> k >> t;
  vector<vector<long long> > dp(2, vector<long long>(200002, 0));
  vector<vector<long long> > sums(2, vector<long long>(200002, 0));
  dp[0][100000] = 1;
  for (int i = 100000; i <= 200000; i++) sums[0][i] = 1;
  for (int i = 1, ii = 0; i <= t; i++, ii = 1 - ii) {
    for (int j = 0; j <= 200000; j++) {
      dp[i % 2][j] = sums[ii][min(j + k, 200000)];
      if (j - k > 0) {
        dp[i % 2][j] -= sums[ii][j - k - 1];
      }
      sums[i % 2][j] = dp[i % 2][j];
      if (j > 0) sums[i % 2][j] += sums[i % 2][j - 1];
      sums[i % 2][j] %= 1000000007;
      dp[i % 2][j] %= 1000000007;
    }
  }
  long long res = 0;
  for (int i = 0; i <= 200000; i++) {
    int pp = i + a - b - 1;
    if (pp < 0) continue;
    if (pp > 200000) pp = 200000;
    res += dp[t % 2][i] * sums[t % 2][pp];
    res %= 1000000007;
  }
  while (res < 0) res += 1000000007;
  cout << res;
}
