http://codeforces.com/problemset/problem/712/D

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

input

1 1 1 2

output

input

2 12 3 1

output

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 are3 + 2 + 1 = 6 possible games in which Memory wins.

题意：两个人最开始有a,b两个初始数，玩t轮游戏，每个人都随机从【-k,k】之间获得一个数加进他们的分数，问最后问最开始是a的人比另外一个大的方案数%1e9就可以；

题解：f[i]表示得分为i的情况数（每轮更新），然后用sum[]来维护f[i]的前缀和，状态转移方程就是f[j]=(sum[loc]-sum[j-k-1]+mod)%mod;（loc=min(j+k,r-k)注意边界）

然后就是数组下表不能小于0；

#include<iostream>
#include<cstdio> 
#include<cstring>
#include<cmath> 
#include<algorithm> 
#define ll long long
double eps=1e-7; 
const int maxn=4e5+2000; 
int zero=2e5+2005; 
const int mod=1e9+7; 
using namespace std;
int f[maxn],sum[maxn]; 
int a,b,k,t; 
int main()
{
    scanf("%d %d %d %d",&a,&b,&k,&t);
    int l=zero-k,r=zero+k; 
    //if(a==1&&b==100&&k==1000&&t==100) 
    for(int i=l;i<=r;i++)
    {
        f[i]=1;
        sum[i]=sum[i-1]+f[i]; 
    } 
    for(int i=2;i<=t;i++)
    {
         l=zero-k*i,r=zero+k*i;
         for(int j=l;j<=r;j++)
         {
             int loc=min(j+k,r-k);
             f[j]=(sum[loc]-sum[j-k-1]+mod)%mod; 
         }
         for(int j=l;j<=r;j++)
         {
             sum[j]=(sum[j-1]+f[j])%mod; 
         } 
    } 
     l=zero-t*k; r=zero+t*k; 
     int ans=0; 
    for(int i=l;i<=r;i++)
    {
        int loc=min(i+a-b-1,r); 
        ans=(ans+((ll)sum[loc]*f[i])%mod)%mod; 
    } 
    printf("%d
",ans); 
     
}