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;
using namespace std;
int mod = 1000 * 1000 * 1000 + 7;
int dp[2][700005];
const int offset = 300000;
int main() {
  int a, b, k, turns;
  cin >> a >> b >> k >> turns;
  turns <<= 1;
  for (int i = offset + 1; i < 700005; ++i) dp[(turns + 1) & 1][i] = 1;
  for (int t = turns; t >= 1; --t) {
    for (int sum = -250001; sum <= 250001; ++sum) {
      dp[(t + 1) & 1][sum + offset] += 1LL * dp[(t + 1) & 1][sum - 1 + offset];
      dp[(t + 1) & 1][sum + offset] %= mod;
    }
    for (int sum = -250001; sum <= 250001; ++sum) {
      int s = sum;
      dp[(t & 1)][s + offset] =
          1LL * dp[(t + 1) & 1][min(700005, s + k + offset)] -
          1LL * dp[(t + 1) & 1][s - k + offset - 1];
      if (dp[t & 1][s + offset] < 0) dp[t & 1][s + offset] += mod;
      dp[t & 1][s + offset] %= mod;
    }
  }
  cout << dp[1][a - b + offset] % mod;
}
