Codeforces Round #370 (Div. 2) D. Memory and Scores DP

D. Memory and Scores

 

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

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：

　　A,B两人玩t轮游戏

　　每轮游戏没人可以从[-k,k]中获取任意的一个分数

　　AB起始分数分别为a,b

　　问你最终A分数严格比B多的方案数

题解：

　　设定dp[i][j]为第i轮 获得分数j的方案数

　　这个可以进行滚动数组和前缀和优化

　　最后枚举一个人的 分数 得到答案

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 5e5+10, M = 1e2+11, mod = 1e9+7, inf = 2e9;

int a,b,k,t;
LL sum[N], dp[2][N];
int main() {
        scanf("%d%d%d%d",&a,&b,&k,&t);
        int now = 1;
        int last = now ^ 1;
        dp[last][0] = 1;
        for(int i = 0; i <= 2 * k * t; ++i) sum[i] = 1;
        for(int i = 1; i <= t; ++i) {
            for(int  j = 0; j <= 2*k*t; ++j) {
                if(j <= 2 * k) dp[now][j] = sum[j];
                else {
                    dp[now][j] = (( sum[j] - sum[j - 2*k - 1] ) % mod + mod ) % mod;
                }
            }
            sum[0] = dp[now][0];
            for(int j = 1; j <= 4 * k * t; ++j)
                sum[j] = ((sum[j-1] + dp[now][j]) % mod + mod) % mod;
            now^=1;
        }
        LL ans = 0;
        for(int i = 0; i <= 2 * k * t; ++i) {
            if(a + i - 1 - b >= 0)ans = (ans + dp[now^1][i] * sum[a + i - 1 - b]%mod) % mod;
        }
        cout<<(ans+mod) % mod<<endl;
        return 0;
}