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;
long long D, K, T;
long long dp[2][10000000];
long long pre[10000000];
long long mi = 0, ma = 0;
long long pl = 0;
long long ans = 0;
long long s = 0;
int32_t main() {
  ios_base::sync_with_stdio(false);
  long long a, b;
  cin >> a >> b;
  D = a - b;
  cin >> K >> T;
  K *= 2;
  mi = D - (K) * (T);
  ma = D + (K) * (T);
  pl = 1000000;
  dp[0][D + pl] = 1;
  for (long long i = 0; i < T; i++) {
    for (long long j = mi; j <= ma; j++) {
      pre[j + pl] = pre[j - 1 + pl] + dp[0][j + pl];
      pre[j + pl] %= 1000000007;
    }
    for (long long j = ma + 1; j <= ma + K + 1; j++) {
      pre[j + pl] = pre[ma + pl];
    }
    s = 0;
    s += (K + 1) * dp[0][mi + pl];
    for (long long j = 1; j <= K; j++) {
      s += (K - j + 1) * dp[0][mi + j + pl];
    }
    s %= 1000000007;
    dp[1][mi + pl] = s;
    for (long long j = mi + 1; j <= ma; j++) {
      long long l = j - K;
      long long r = j + K;
      s -= (pre[j - 1 + pl] - pre[l - 2 + pl]);
      s += (pre[r + pl] - pre[j - 1 + pl]);
      s %= 1000000007;
      s += 1000000007;
      s %= 1000000007;
      dp[1][j + pl] = s;
    }
    for (long long j = mi; j <= ma; j++) {
      dp[0][j + pl] = dp[1][j + pl];
    }
  }
  for (long long j = 1; j <= ma; j++) {
    ans += dp[0][j + pl];
    ans %= 1000000007;
  }
  cout << ans << '\n';
  return 0;
}
