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;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
const int INF = 0x3f3f3f3f, N = 4.1e5 + 5;
const ll LINF = 1e18 + 5;
constexpr int mod = 1e9 + 7;
int a, b, k, t;
ll dp[2][N], sum[2][N], g[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> a >> b >> k >> t;
  for (int i = -2 * k, j = 1; i <= 0; i++, j++) g[i + 2 * k] = j;
  for (int i = 1, j = 2 * k; i <= 2 * k; i++, j--) g[i + 2 * k] = j;
  const int M = 2e5 + 200 + a - b;
  int cur = 1;
  dp[0][M] = 1;
  for (int i = 0; i <= 4 * k; i++) sum[0][i + M] = 1;
  auto f = [&](int x, int l, int r) {
    return (sum[x][r] - sum[x][l - 1] + mod) % mod;
  };
  for (int i = 1; i <= t; i++) {
    ll pre = 0;
    for (int j = -2 * i * k, w = -2 * k; w <= 2 * k; w++)
      (pre += dp[cur ^ 1][j + w + M] * g[-w + 2 * k] % mod) %= mod;
    for (int j = -2 * i * k; j <= 2 * i * k; j++) {
      dp[cur][j + M] = pre;
      pre = (pre - f(cur ^ 1, j - 2 * k + M, j + M) +
             f(cur ^ 1, j + 1 + M, j + 2 * k + 1 + M)) %
            mod;
      pre = (pre + mod) % mod;
    }
    sum[cur][-2 * (i + 2) * k + M] = dp[cur][-2 * (i + 2) * k + M];
    for (int j = -2 * (i + 2) * k + M + 1; j <= 2 * (i + 2) * k + M; j++)
      (sum[cur][j] = sum[cur][j - 1] + dp[cur][j]) %= mod;
    cur ^= 1;
  }
  ll ans = 0;
  for (int i = -2 * t * k; i <= 2 * t * k; i++)
    if (i + a - b > 0) ans = (ans + dp[cur ^ 1][i + M]) % mod;
  cout << ans;
  return 0;
}
