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>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
string to_string(char c) { return "'" + string(1, c) + "'"; }
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ": " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = false;
  string r = "{";
  for (auto x : v) {
    if (f) r += ", ";
    r += to_string(x);
    f = true;
  }
  return r += "}";
}
template <typename A>
string to_string(vector<vector<A>> v) {
  string r;
  for (auto x : v) r += "\n" + to_string(x);
  return r;
}
int Nerr;
template <typename A>
string to_string(A *p) {
  return to_string(vector<A>(p, p + Nerr));
}
void err(istream_iterator<string>) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " =: " << to_string(a) << "; ";
  err(++it, args...);
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
int const MOD = 1e9 + 7;
long long const INF = 1e18 + 42;
inline int add(int a, int b, int mod = MOD) {
  a += b;
  return a >= mod ? a - mod : a;
}
inline int sub(int a, int b, int mod = MOD) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int mul(int a, int b, int mod = MOD) {
  return int((long long)a * b % mod);
}
inline int mpow(int base, long long ex, int mod = MOD) {
  int res = 1;
  for (; ex > 0; ex >>= 1) {
    if (ex & 1) res = mul(res, base, mod);
    base = mul(base, base, mod);
  }
  return res;
}
inline int inv(int a, int mod = MOD) { return mpow(a, mod - 2, mod); }
inline int mdiv(int a, int b, int mod = MOD) {
  return mul(a, mpow(b, mod - 2, mod));
}
inline void adds(int &a, int b, int mod = MOD) {
  a += b;
  if (a >= mod) a -= mod;
}
inline void subs(int &a, int b, int mod = MOD) {
  a -= b;
  if (a < 0) a += mod;
}
inline void muls(int &a, int b, int mod = MOD) {
  a = int((long long)a * b % mod);
}
inline void mdivs(int &a, int b, int mod = MOD) { a = mdiv(a, b, mod); }
const int N = 5e5 + 42;
const int M = 2.5e5 + 12;
int dp[N];
int a[N];
int b[N];
void clear() {
  memset(a, 0, sizeof a);
  memset(b, 0, sizeof b);
}
void upd(int l, int r, int s, int d) {
  if (l < 0) {
    adds(s, mul(-l, d));
    l = 0;
  }
  uin(r, N - 2);
  if (r < l) return;
  adds(a[l], s);
  subs(a[r + 1], add(s, mul(r - l, d)));
  adds(b[l + 1], d);
  subs(b[r + 1], d);
}
void recalc() {
  int cur = 0, d = 0;
  for (int i = 0; i < N; ++i) {
    adds(d, b[i]);
    adds(cur, a[i]);
    adds(cur, d);
    dp[i] = cur;
  }
}
int32_t main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int A, B, k, t;
  cin >> A >> B >> k >> t;
  dp[A - B + M] = 1;
  for (int i = 0; i < t; ++i) {
    clear();
    for (int j = 0; j < N; ++j)
      if (dp[j]) {
        upd(j, j + 2 * k, mul(dp[j], 2 * k + 1), MOD - dp[j]);
        upd(j - 2 * k, j - 1, dp[j], dp[j]);
      }
    recalc();
  }
  int ans = 0;
  for (int i = M + 1; i < N; ++i) {
    adds(ans, dp[i]);
  }
  kek(ans);
}
