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 long long MOD = 1e9 + 7, MAX = 5e5 + 5, ZERO = MAX / 2;
int a, b, k, t;
long long dp[2][MAX];
long long pref[MAX];
inline int getidx(int x) { return x + ZERO; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> a >> b >> k >> t;
  for (int i = 0; i < (int)(MAX); i++) dp[0][i] = 0;
  dp[0][getidx(a - b)] = 1;
  for (int turn = (1); turn <= (int)(t); turn += (1)) {
    long long sum = 0;
    for (int i = (0); i <= (int)(2 * k); i += (1))
      sum += dp[~turn & 1][i] * (i + 1) % MOD, sum %= MOD;
    for (int i = (2 * k + 1); i <= (int)(4 * k); i += (1))
      sum += dp[~turn & 1][i] * (4 * k - i + 1) % MOD, sum %= MOD;
    for (int i = 0; i < (int)(MAX); i++) {
      pref[i] = dp[~turn & 1][i];
      if (i) pref[i] += pref[i - 1], pref[i] %= MOD;
    }
    dp[turn & 1][2 * k] = sum;
    for (int i = (2 * k + 1); i <= (int)(MAX - 2 * k - 1); i += (1)) {
      long long minus =
          (pref[i - 1] +
           (MOD - (i >= 2 * k + 2 ? pref[i - 2 * k - 2] : 0)) % MOD) %
          MOD;
      long long plus = (pref[i + 2 * k] + (MOD - pref[i - 1]) % MOD) % MOD;
      sum += (MOD - minus) % MOD, sum %= MOD;
      sum += plus, sum %= MOD;
      dp[turn & 1][i] = sum;
    }
  }
  long long ans = 0;
  for (int i = (getidx(1)); i <= (int)(MAX - 1); i += (1)) {
    ans += dp[t & 1][i], ans %= MOD;
  }
  cout << ans << '\n';
  return 0;
}
