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 MOD = 1e9 + 7;
int main() {
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  t <<= 1;
  long long dp[2 * t * k + 1];
  long long pref[2 * t * k + 1];
  memset(dp, 0, sizeof dp);
  memset(pref, 0, sizeof pref);
  int offset = t * k;
  for (int i = -k; i <= k; ++i) {
    dp[offset + i] = 1;
  }
  for (int j = 1; j <= 2 * k * t; ++j) pref[j] = (dp[j] + pref[j - 1]) % MOD;
  for (int i = 1; i < t; ++i) {
    for (int j = -(k * (i + 1)); j <= (k * (i + 1)); ++j) {
      if (offset + j + k > 2 * k * t)
        dp[offset + j] = pref[2 * k * t];
      else
        dp[offset + j] = pref[offset + j + k];
      if (offset + j - k - 1 >= 0) dp[offset + j] -= pref[offset + j - k - 1];
      dp[offset + j] = (dp[offset + j] % MOD + MOD) % MOD;
    }
    pref[0] = dp[0];
    for (int j = 1; j <= 2 * k * t; ++j) pref[j] = (dp[j] + pref[j - 1]) % MOD;
  }
  long long ans = 0;
  for (int i = offset + b - a + 1; i <= 2 * k * t; ++i) {
    ans += dp[i];
    ans %= MOD;
  }
  cout << ans << endl;
}
