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>
const char en = '\n';
using namespace std;
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U>& par) {
  out << par.first << " " << par.second << en;
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (long long i = 0; i < (long long)(v.size()); i++) {
    if (i) out << " ";
    cout << v[i];
  }
  return out;
}
long long prv = 1000000007;
vector<long long> f(long long a, long long k, long long t) {
  vector<long long> V(300000, 0);
  V[110000 + a] = 1;
  for (long long i = 0; i < (long long)(t); i++) {
    long long sum = 0;
    vector<long long> X(300000, 0);
    for (long long j = (long long)(1005); j <= (long long)(230000); j++) {
      sum += V[j + k];
      sum -= V[j - k - 1];
      sum += 3 * prv;
      sum %= prv;
      X[j] += sum;
      X[j] %= prv;
    }
    swap(X, V);
  }
  return V;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  vector<long long> A = f(a, k, t);
  vector<long long> B = f(b, k, t);
  vector<long long> P = B;
  long long suc = 0, r = 0;
  for (long long i = 0; i < (long long)(230000); i++) {
    r += suc * A[i];
    suc += B[i];
    r %= prv;
    suc %= prv;
  }
  cout << r << endl;
}
