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 double PI = 3.141592653589;
template <class T>
void dbs(string str, T t) {
  cerr << str << " : " << t << "\n";
}
template <class T, class... S>
void dbs(string str, T t, S... s) {
  long long idx = str.find(',');
  cerr << str.substr(0, idx) << " : " << t << ", ";
  dbs(str.substr(idx + 1), s...);
}
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream& operator<<(ostream& os, const map<S, T>& p) {
  os << "[ ";
  for (auto& it : p) os << it << " ";
  return os << "]";
}
template <class T>
void prc(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}
long long a, b, k, t;
const long long MAX = 1e5 + 1e2 + 5;
long long ipow(long long base, long long exp) {
  long long ans = 1;
  while (exp > 0) {
    if (exp & 1) ans = (ans * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp >>= 1;
  }
  return ans;
}
long long fact[2 * MAX + 1];
long long inv_fact[2 * MAX + 1];
long long nck(long long n, long long k) {
  if (k > n) return 0;
  long long ans = (fact[n] * inv_fact[k]) % 1000000007;
  ans = (ans * inv_fact[n - k]) % 1000000007;
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> k >> t;
  fact[0] = 1;
  for (long long i = 1; i <= 2 * MAX; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
  inv_fact[2 * MAX] = ipow(fact[2 * MAX], 1000000007 - 2);
  for (long long i = 2 * MAX - 1; i > 0; i--)
    inv_fact[i] = (inv_fact[i + 1] * (i + 1)) % 1000000007;
  inv_fact[0] = 1;
  vector<long long> dp1(2 * MAX + 1, 0), dp2(2 * MAX + 1, 0);
  for (long long i = a; i <= a + 2 * k * t; i++) {
    dp1[i] = nck(i + t - 1 - a, t - 1);
    long long neg = 0, sign = 1;
    for (long long j = 1; j <= t; j++) {
      neg = (neg +
             nck(t, j) * nck(i + t - 1 - a - (2 * k + 1) * j, t - 1) * sign) %
            1000000007;
      sign *= -1;
    }
    dp1[i] = (dp1[i] - neg + 1000000007) % 1000000007;
  }
  for (long long i = b; i <= b + 2 * k * t; i++) {
    dp2[i] = nck(i + t - 1 - b, t - 1);
    long long neg = 0, sign = 1;
    for (long long j = 1; j <= t; j++) {
      neg = (neg +
             nck(t, j) * nck(i + t - 1 - b - (2 * k + 1) * j, t - 1) * sign) %
            1000000007;
      sign *= -1;
    }
    dp2[i] = (dp2[i] - neg + 1000000007) % 1000000007;
  }
  for (long long i = 1; i <= 2 * MAX; i++)
    dp2[i] = (dp2[i] + dp2[i - 1]) % 1000000007;
  long long ans = 0;
  for (long long i = 1; i <= 2 * MAX; i++) {
    ans = (ans + dp1[i] * dp2[i - 1]) % 1000000007;
  }
  cout << ans << "\n";
  return 0;
}
