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 N = (long long)(5e5) + 322;
const long long INF = (long long)(1e9);
const long long mod = (long long)(1e9) + 7;
const double eps = 1e-9;
long long a, b, t, k, pref[N], dp[N], mid, L = 10000, R = 470000, ans;
inline long long get(long long l, long long r) {
  return (pref[r] - pref[l - 1] + mod) % mod;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> a >> b >> k >> t;
  k *= 2;
  mid = 250000;
  dp[a - b + mid] = 1;
  for (long long i = 1; i <= N - 100; ++i)
    pref[i] = (dp[i] + pref[i - 1]) % mod;
  for (long long j = 1; j <= t; ++j) {
    for (long long i = L; i <= R; ++i)
      dp[i] = (dp[i - 1] - get(i - k - 1, i - 1) + mod + get(i, i + k)) % mod;
    for (long long i = 1; i <= N - 100; ++i)
      pref[i] = (pref[i - 1] + dp[i]) % mod;
  }
  for (long long i = mid + 1; i <= R; ++i) ans += dp[i], ans %= mod;
  cout << ans;
  return 0;
}
