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>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const int LIM = 8e5 + 30, MOD = 1e9 + 7;
int t, n, m, k, x, y;
int o = 4e5 + 10;
long long int diff[5][(int)8e5 + 300] = {0};
void add(int i, int l, int u, long long int v) {
  if (v == 0) return;
  assert(u + o + 1 < LIM);
  assert(l + o >= 0);
  diff[i][l + o] = (diff[i][l + o] + v) % MOD;
  diff[i][u + o + 1] = (diff[i][u + o + 1] + MOD - v) % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  add(0, a - b, a - b, 1);
  int ind = 0;
  for (int i = 0; i < 2 * t; i++) {
    long long int val = 0;
    int nxtind = (ind + 1) % 5;
    bool st = 0;
    for (int curr = 0; curr < LIM; curr++) {
      val += diff[ind][curr];
      diff[ind][curr] = 0;
      val %= MOD;
      add(nxtind, curr - o - k, curr - o + k, val);
      if (val) st = 1;
      if (st && !val) break;
    }
    ind = nxtind;
  }
  long long int ans = 0, curr = 0;
  for (int i = 0; i < LIM; i++) {
    curr += diff[ind][i];
    curr %= MOD;
    if (i > o) {
      ans += curr, ans %= MOD;
    }
  }
  cout << ans << "\n";
  return 0;
}
