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 = 1e9 + 7;
const int OFFSET = 100000;
long long dp[2 * 100005][2];
long long sum[2 * 100005][2];
int k, t;
long long get_sum(int left, int right, int flag) {
  if (right >= k * t) right = k * t;
  bool output = false;
  long long ans = sum[right + OFFSET][flag];
  left = left + OFFSET - 1;
  if (left >= 0) ans -= sum[left][flag];
  if (ans < 0) ans += mod;
  ans %= mod;
  return ans;
}
int main() {
  int a, b;
  cin >> a >> b >> k >> t;
  for (int i = -k; i <= k; i++) {
    dp[i + OFFSET][1] = 1;
    sum[i + OFFSET][1] = 1 + sum[i + OFFSET - 1][1];
  }
  for (int i = k + 1; i <= 100000; i++)
    sum[i + OFFSET][1] = sum[i + OFFSET - 1][1];
  for (int i = 2; i <= t; i++) {
    int flag = i & 1;
    for (int j = -k * t; j <= k * t; j++) {
      dp[j + OFFSET][flag] = get_sum(j - k, j + k, flag ^ 1);
      dp[j + OFFSET][flag] %= mod;
    }
    sum[-k * t + OFFSET][flag] = dp[-k * t + OFFSET][flag];
    for (int j = -k * t + 1; j <= k * t; j++) {
      sum[j + OFFSET][flag] = dp[j + OFFSET][flag] + sum[j + OFFSET - 1][flag];
      sum[j + OFFSET][flag] %= mod;
    }
  }
  long long ans = 0;
  for (int i = k * t + 1; i <= 100000; i++)
    sum[i + OFFSET][t & 1] = sum[i + OFFSET - 1][t & 1];
  for (int i = -k * t; i <= k * t; i++) {
    int na = a + i;
    int max_n = na - b - 1 + OFFSET;
    if (max_n >= 0) {
      long long to =
          dp[i + OFFSET][t & 1] * sum[min(k * t + OFFSET, max_n)][t & 1];
      to %= mod;
      ans += to;
      ans %= mod;
    }
  }
  cout << ans << endl;
  return 0;
}
