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 MAX = 1000005;
const int MOD = 1000000007;
int f[MAX];
int fi[MAX];
int pos(int a) { return ((long long)a % MOD + MOD) % MOD; }
int add(int a, int b) { return ((long long)a + (long long)b) % MOD; }
int sub(int a, int b) { return pos(a - b); }
int mult(int a, int b) { return (long long)pos(a) * b % MOD; }
int sq(int a) { return (long long)a * a % MOD; }
int expo(int a, int b) {
  if (b == 0) {
    return 1;
  }
  if (b % 2) {
    return mult(a, sq(expo(a, b / 2)));
  } else {
    return sq(expo(a, b / 2));
  }
}
int inv(int a) { return expo(a, MOD - 2); }
int c(int n, int k) { return mult(f[n], mult(fi[k], fi[n - k])); }
int poly1[MAX];
int pref2[MAX];
int main() {
  f[0] = fi[0] = 1;
  for (int i = 1; i < MAX; i++) {
    f[i] = mult(f[i - 1], i);
    fi[i] = inv(f[i]);
  }
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  if (b - a > 2 * k * t) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 2 * t + 1; i++) {
    poly1[(2 * k + 1) * i] = ((i % 2 == 0) ? 1 : -1) * c(2 * t, i);
  }
  for (int i = 0; i < MAX; i++) {
    pref2[i] = c(2 * t + i, 2 * t);
  }
  int lb = 2 * k * t + b - a + 1;
  int ub = 4 * k * t;
  int ans = 0;
  bool f = true;
  for (int i = 0; i <= 2 * t, f == true; i++) {
    int l = lb - (2 * k + 1) * i;
    int u = ub - (2 * k + 1) * i;
    if (u < 0) {
      f = false;
    }
    if (l < 0) {
      ans = add(ans, mult(poly1[(2 * k + 1) * i], pref2[u]));
    } else {
      ans = add(ans, mult(poly1[(2 * k + 1) * i], sub(pref2[u], pref2[l - 1])));
    }
  }
  cout << ans << endl;
}
