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>
double eps = 1e-9;
using namespace std;
int c = 100000;
long long dp[2][300005], DP[2][300005], sum[300005], a, b, k, t;
int main() {
  cin >> a >> b >> k >> t;
  dp[0][a + c] = 1;
  for (long long i = a + c; i <= 220000; i++) sum[i] = 1;
  for (long long i = 1; i <= t; i++) {
    for (long long j = a - k * t; j <= a + t * k; j++) {
      int want = j + c;
      int low = want - k, high = want + k;
      low = (low > 1 ? low : 1);
      high = (high < 220000 ? high : 220000);
      dp[i % 2][want] = sum[high] - sum[low - 1];
      dp[i % 2][want] += 1000000007;
      dp[i % 2][want] %= 1000000007;
    }
    memset(sum, 0, sizeof sum);
    memset(dp[((i % 2) + 1) % 2], 0, sizeof dp[((i % 2) + 1) % 2]);
    for (long long j = a - t * k; j <= a + t * k; j++) {
      int want = j + c;
      sum[want] = (dp[i % 2][want] + sum[want - 1]) % 1000000007;
    }
    for (long long j = a + t * k + c + 1; j <= 220000; j++) sum[j] = sum[j - 1];
  }
  DP[0][b + c] = 1;
  memset(sum, 0, sizeof sum);
  for (long long i = b + c; i <= 220000; i++) sum[i] = 1;
  for (long long i = 1; i <= t; i++) {
    for (long long j = b - t * k; j <= b + t * k; j++) {
      int want = j + c;
      int low = want - k, high = want + k;
      low = (low > 1 ? low : 1);
      high = (high < 220000 ? high : 220000);
      DP[i % 2][want] = sum[high] - sum[low - 1];
      DP[i % 2][want] += 1000000007;
      DP[i % 2][want] %= 1000000007;
    }
    memset(sum, 0, sizeof sum);
    memset(DP[((i % 2) + 1) % 2], 0, sizeof DP[((i % 2) + 1) % 2]);
    for (long long j = b - t * k; j <= b + t * k; j++) {
      int want = j + c;
      sum[want] = (DP[i % 2][want] + sum[want - 1]) % 1000000007;
    }
    for (long long j = b + t * k + c + 1; j <= 220000; j++) sum[j] = sum[j - 1];
  }
  long long ans = 0;
  for (long long i = a - t * k; i <= a + t * k; i++) {
    int want = i + c;
    ans = (ans + (dp[t % 2][want] * sum[want - 1]) % 1000000007) % 1000000007;
  }
  cout << ans << endl;
}
