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 GCC optimize("O3")
#pragma GCC target("tune=native")
using namespace std;
const int MAX = 9e5 + 5;
const long long MOD = 1000000007;
const long long MOD2 = 2010405347;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 600;
int a, b, k, t, le, ri, x, y, dp[2][MAX];
long long ans;
bool ls, nw;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> k >> t;
  t <<= 1;
  le = ri = 450000 + a - b;
  dp[1][le] = 1;
  ls = 1;
  for (int iter = 1; iter <= t; ++iter) {
    memset(dp[nw], 0, sizeof dp[nw]);
    for (int i = le; i <= ri; ++i)
      dp[nw][i - k] += dp[ls][i], dp[nw][i + k + 1] -= dp[ls][i];
    le -= k, ri += k;
    for (int i = le + 1; i <= ri; ++i) {
      dp[nw][i] += dp[nw][i - 1];
      if (dp[nw][i] < 0)
        dp[nw][i] += MOD;
      else if (dp[nw][i] >= MOD)
        dp[nw][i] -= MOD;
    }
    nw ^= 1, ls ^= 1;
  }
  for (int i = 450001; i <= ri; ++i) ans += dp[ls][i];
  cout << ans % MOD << '\n';
  return 0;
}
