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 int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int mod = 1000000007;
const double pi = acos(-1.0);
inline void gn(long long &x) {
  int sg = 1;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  c == '-' ? (sg = -1, x = 0) : (x = c - '0');
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= sg;
}
inline void gn(int &x) {
  long long t;
  gn(t);
  x = t;
}
inline void gn(unsigned long long &x) {
  long long t;
  gn(t);
  x = t;
}
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
long long powmod(long long a, long long x, long long mod) {
  long long t = 1ll;
  while (x) {
    if (x & 1) t = t * a % mod;
    a = a * a % mod;
    x >>= 1;
  }
  return t;
}
const int maxn = 2000 * 100 * 2 * 2;
int dp[maxn], pd[maxn];
inline void add(int &a, int b) {
  a += b;
  if (a > mod) a -= mod;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int main() {
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  int mx = 4 * k * t;
  dp[k * 2 * t] = 1;
  t *= 2;
  for (int _ = 1; _ <= t; ++_) {
    memset(pd, 0, sizeof(pd));
    int l = -k, r = k;
    int su = 0;
    for (int i = 0; i <= r; ++i) add(su, dp[i]);
    for (int i = 0; i <= mx; ++i) {
      pd[i] = su;
      if (l >= 0) sub(su, dp[l]);
      l++;
      r++;
      add(su, dp[r]);
    }
    for (int i = 0; i <= mx; ++i) dp[i] = pd[i];
  }
  int ans = 0;
  for (int i = 0; i <= mx; ++i)
    if (i > b - a + t * k) add(ans, dp[i]);
  cout << ans << endl;
  return 0;
}
