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 N = 2 * 1e5 + 1;
long long dp[101][N], a[N], b[N];
long long x, y, k, t;
const long long mod = 1e9 + 7;
int main() {
  cin >> x >> y >> k >> t;
  long long up = k * t * 2;
  dp[0][k * t] = 1;
  for (long long i = 0; i <= up; i++) {
    if (i == 0)
      a[i] = dp[0][i];
    else
      a[i] = a[i - 1] + dp[0][i];
  }
  long long *p = a, *q = b, *g;
  for (long long i = 1; i <= t; i++) {
    for (long long j = 0; j <= up; j++) {
      long long l = j - k, r = j + k;
      l = max((long long)0, l);
      r = min(up, r);
      long long t = dp[i][j];
      if (l == 0)
        dp[i][j] = (dp[i][j] + p[r]) % mod;
      else
        dp[i][j] = (dp[i][j] + p[r] - p[l - 1] + mod) % mod;
      if (j == 0)
        q[j] = dp[i][j] % mod;
      else
        q[j] = (dp[i][j] + q[j - 1]) % mod;
    }
    g = q;
    q = p;
    p = g;
  }
  long long ans = 0;
  long long dis = y - x + 1;
  for (long long i = 0; i <= up; i++) {
    long long now = i - dis;
    if (now > up) now = up;
    if (now >= 0) {
      ans = (ans + ((dp[t][i]) * p[now]) % mod) % mod;
    }
  }
  cout << ans % mod << endl;
  return 0;
}
