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, maxn = 420000;
int a, b, k, t, dp[2][maxn], sum[maxn];
int n = 400000, m = 200000;
void add(int &x, int y) {
  x = x + y >= 1000000007 ? x + y - 1000000007 : x + y;
}
int Solve(int l, int r) {
  l = max(1, l), r = min(r, n);
  return (sum[r] - sum[l - 1] + 1000000007) % 1000000007;
}
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  int cur = 0;
  dp[cur][m] = 1;
  for (int i = 1; i <= t; i++) {
    cur ^= 1;
    memset(dp[cur], 0, sizeof(dp[cur]));
    sum[0] = dp[cur ^ 1][0];
    for (int j = 1; j <= n; j++)
      sum[j] = (sum[j - 1] + dp[cur ^ 1][j]) % 1000000007;
    for (int j = 0; j <= n; j++) {
      if (j == 0) {
        for (int l = 0; l <= 2 * k; l++)
          add(dp[cur][j],
              (long long)(2 * k + 1 - l) * dp[cur ^ 1][j + l] % 1000000007);
      } else {
        dp[cur][j] = dp[cur][j - 1];
        add(dp[cur][j], 1000000007 - Solve(j - 2 * k - 1, j - 1));
        add(dp[cur][j], Solve(j, j + 2 * k));
      }
    }
  }
  int ans = 0;
  for (int i = m + 1 + b - a; i <= n; i++) add(ans, dp[cur][i]);
  printf("%d\n", ans);
  return 0;
}
