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;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &x) {
  s << "[";
  for (auto it : x) {
    s << it << ", ";
  }
  s << "]";
  return s;
}
template <typename T>
ostream &operator<<(ostream &s, const set<T> &x) {
  s << "{";
  for (auto it : x) {
    s << it << ", ";
  }
  s << "}";
  return s;
}
template <typename U, typename V>
ostream &operator<<(ostream &s, const pair<U, V> &x) {
  s << "(" << x.first << ", " << x.second << ")";
  return s;
}
template <typename T>
bool chmax(T &x, const T &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T &x, const T &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
const int maxn = 3e5;
int a[maxn], sum[maxn];
const long long MOD = 1e9 + 7;
int main() {
  int A, B, k, t;
  cin >> A >> B >> k >> t;
  k = 2 * k + 1;
  a[0] = 1;
  for (int i = 0; i < maxn; i++) {
    sum[i] = 1;
  }
  while (t--) {
    for (int i = 0; i < maxn; i++) {
      a[i] = sum[i];
      if (i >= k) {
        a[i] -= sum[i - k];
      }
      if (a[i] < 0) {
        a[i] += MOD;
      }
    }
    for (int i = 0; i < maxn; i++) {
      if (i == 0) {
        sum[i] = a[i];
      } else {
        sum[i] = a[i] + sum[i - 1];
        if (sum[i] >= MOD) {
          sum[i] -= MOD;
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < maxn; i++) {
    int j = i - B + A;
    j--;
    if (j >= 0 && j < maxn) {
      ans = (ans + a[i] * 1ll * sum[j]) % MOD;
    }
  }
  cout << ans << "\n";
  return 0;
}
