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;
long long a, b, k, t;
long long dp[105][2 * 1005 * 105];
long long prod;
long long adj(long long x) { return x + prod; }
int main() {
  cin >> a >> b >> k >> t;
  prod = t * k;
  for (long long i = -k; i <= k; i++) {
    dp[1][adj(i)] = i + k + 1;
  }
  for (long long i = k + 1; i <= prod; i++) {
    dp[1][adj(i)] = dp[1][adj(i - 1)];
  }
  for (long long i = 2; i <= t; i++) {
    for (long long j = -prod; j <= prod; j++) {
      long long endval = dp[i - 1][adj(prod)];
      long long begval = 0;
      if (j - k > -prod) begval = dp[i - 1][adj(j - k) - 1];
      if (j + k <= prod) endval = dp[i - 1][adj(j + k)];
      dp[i][adj(j)] = (dp[i][adj(j) - 1] + endval - begval) % 1000000007;
    }
  }
  long long res = 0;
  for (long long i = -prod; i <= prod; i++) {
    long long ways = dp[t][adj(i)] - dp[t][adj(i) - 1];
    long long x = b + i - a + 1;
    if (x < -prod) {
      res = (res + ways * (dp[t][adj(prod)])) % 1000000007;
    } else if (x > prod) {
      break;
    } else {
      res = (res + ways * (dp[t][adj(prod)] - dp[t][adj(x) - 1])) % 1000000007;
    }
  }
  cout << (1000000007 + res) % 1000000007 << endl;
  return 0;
}
