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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-8;
const long long mod = 1e9 + 7;
const unsigned long long mx = 133333331;
inline void RI(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0';
}
long long dp[2][400005];
long long sum[400005];
int main() {
  int a, b, k, t;
  while (cin >> a >> b >> k >> t) {
    int cnt = 0;
    memset(dp, 0, sizeof(dp));
    dp[cnt ^ 1][200000] = 1;
    for (int i = 1; i <= t; i++) {
      sum[0] = dp[cnt ^ 1][0];
      memset(dp[cnt], 0, sizeof(dp[cnt]));
      for (int j = 1; j <= 400000; j++)
        sum[j] = (sum[j - 1] + dp[cnt ^ 1][j]) % mod;
      for (int j = 0; j <= 400000; j++) {
        if (j == 0) {
          dp[cnt][j] += (2 * k + 1) * dp[cnt ^ 1][j] % mod;
          for (int z = 1; z <= 2 * k; z++)
            dp[cnt][j] += (2 * k + 1 - z) * dp[cnt ^ 1][j + z] % mod;
        } else {
          dp[cnt][j] = dp[cnt][j - 1];
          if (j - 2 * k < 2)
            dp[cnt][j] -= sum[j - 1];
          else
            dp[cnt][j] -= sum[j - 1] - sum[j - 2 - 2 * k];
          if (j + 2 * k > 400000)
            dp[cnt][j] += sum[400000] - sum[j - 1];
          else
            dp[cnt][j] += sum[j + 2 * k] - sum[j - 1];
        }
        dp[cnt][j] = (dp[cnt][j] % mod + mod) % mod;
      }
      cnt ^= 1;
    }
    long long ans = 0;
    for (int i = 200001 - (a - b); i <= 400000; i++)
      ans = (ans + dp[cnt ^ 1][i]) % mod;
    cout << ans << endl;
  }
  return 0;
}
