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;
void read(int &x) {
  bool neg = false;
  register int c;
  x = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  if (neg) x *= -1;
}
long long sub(long long a, long long b) {
  long long res = (a - b) % 1000000007;
  if (res < 0) res += 1000000007;
  return res;
}
int dp[102][200110], dp2[102][200110];
int main() {
  {
    long long a, b, K, T;
    scanf("%lld", &a), scanf("%lld", &b), scanf("%lld", &K), scanf("%lld", &T);
    K = K * 2;
    dp[0][a] = 1;
    for (long long k = 1; k <= K * T + a; k++) dp[0][k] += dp[0][k - 1];
    for (int i = 1; i <= T; i++) {
      for (long long k = 0; k <= K * T + a; k++) {
        dp[i][k] =
            (dp[i - 1][k] - (k - K > 0 ? dp[i - 1][max(0LL, k - K - 1)] : 0)) %
            1000000007;
        if (dp[i][k] < 0) dp[i][k] += 1000000007;
        if (k > 0) dp[i][k] = (dp[i][k] + dp[i][k - 1]) % 1000000007;
      }
    }
    dp2[0][b] = 1;
    for (long long k = 1; k <= K * T + b; k++) dp2[0][k] += dp2[0][k - 1];
    for (int i = 1; i <= T; i++) {
      for (long long k = 0; k <= K * T + b; k++) {
        dp2[i][k] = (dp2[i - 1][k] -
                     (k - K > 0 ? dp2[i - 1][max(0LL, k - K - 1)] : 0)) %
                    1000000007;
        if (dp2[i][k] < 0) dp2[i][k] += 1000000007;
        if (k > 0) dp2[i][k] = (dp2[i][k] + dp2[i][k - 1]) % 1000000007;
      }
    }
    long long ans = 0;
    for (long long i = 0; i <= K * T + b; i++) {
      if (i <= K * T + a)
        ans =
            (ans % 1000000007 + (sub(dp2[T][i], i > 0 ? dp2[T][i - 1] : 0) *
                                 sub(dp[T][K * T + a], i >= 0 ? dp[T][i] : 0)) %
                                    1000000007) %
            1000000007;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
