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 dp[105][430005], cf[500005], cfcf[500005];
int qwq = 207000, mx = 420000;
int main(int argc, char** argv) {
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  dp[0][a - b + qwq] = 1;
  for (int i = 0; i <= t; i++) {
    for (int j = 1; j <= mx; j++)
      cfcf[j] += cfcf[j - 1], cfcf[j] %= 1000000007, cf[j] += cfcf[j],
          cf[j] %= 1000000007;
    for (int j = 1; j <= mx; j++)
      cf[j] += cf[j - 1], cf[j] %= 1000000007, dp[i][j] += cf[j],
          dp[i][j] %= 1000000007;
    memset(cf, 0, sizeof cf), memset(cfcf, 0, sizeof cfcf);
    for (int j = k + k; j <= mx; j++)
      cfcf[j - k - k] += dp[i][j], cfcf[j - k - k] %= 1000000007,
          cfcf[j + 1] -= (dp[i][j] + dp[i][j]) % 1000000007,
          cfcf[j + 1] %= 1000000007, cfcf[j + k + k + 2] += dp[i][j],
          cfcf[j + k + k + 2] %= 1000000007;
  }
  long long ans = 0;
  for (int i = qwq + 1; i <= mx; i++) ans += dp[t][i];
  cout << (ans % 1000000007 + 1000000007) % 1000000007;
  return 0;
}
