Memory and Scores

Memory and Scores

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

分析：dp[i][j]表示第i轮获得分数为j的方案数；

　　　计数时维护前缀和即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t;
ll dp[2][maxn],sum[maxn],a,b;
int main()
{
    int i,j;
    scanf("%lld%lld%d%d",&a,&b,&k,&t);
    dp[0][0]=1;
    int now=1;
    rep(i,0,2*k*t)sum[i]=1;
    rep(i,1,t)
    {
        rep(j,0,2*k*t)
        {
            if(j<=2*k)dp[now][j]=sum[j];
            else dp[now][j]=(sum[j]-sum[j-2*k-1]+mod)%mod;
        }
        sum[0]=dp[now][0];
        rep(j,1,2*k*t+100)
        {
            sum[j]=(sum[j-1]+dp[now][j])%mod;
        }
        now^=1;
    }
    ll ans=0;
    rep(i,0,2*k*t)
    {
        if(i+a-b-1>=0)ans=(ans+dp[now^1][i]*sum[i+a-b-1]%mod)%mod;
    }
    printf("%lld
",ans);
    //system("Pause");
    return 0;
}