codeforces 712D. Memory and Scores（dp）

D. 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 are 3 + 2 + 1 = 6 possible games in which Memory wins.

题意：

Memory跟Lexa玩游戏，起始分数分别是a、b，他们玩了t局，每玩一局，他们可以获得一个随机分数，分数值范围[-k,k]，求最后Memory的分数比Lexa大的结果有几种。

思路：

dp[i][j]表示玩到第i次游戏时，得分是j的情况。

dp[i][j] = ∑dp[i-1][l], l∈[-k, k]

当Memory的分数是x时，则Lexa的分数y∈[-t*k, x+a-b-1]，一共有dp[t][x]*dp[t][y]种情况

枚举Memory的分数，求和

【我的代码中注释掉的部分是暴力的求解方式，会超时，要求前缀和优化】

#include <bits/stdc++.h>
using namespace std;
#define ll __int64
const int N=5e5+10;
const int shift=1e5+100;
const ll mod = 1e9+7;
ll dp[110][N];
ll pre[2][N];

int main(){
	int a, b, k, t;
	scanf("%d%d%d%d", &a, &b, &k, &t);
	
	dp[0][shift] = 1;
	int now=0;
	pre[now][shift]=1;
	for(int i=1; i<=t; i++){
		pre[!now][-i*k+shift-1]=0;
		for(int j=-i*k; j<=i*k; j++){
			/*for(int l=-k; l<=k; l++){
				dp[i][j+shift] = (dp[i][j+shift]+dp[i-1][j-l+shift])%mod;
			}*/
			int left = j-k-1, right = j+k;
			ll s2 = left<-(i-1)*k ? 0:pre[now][left+shift];
			ll s1 = right>(i-1)*k ? pre[now][(i-1)*k+shift]:pre[now][right+shift];
			
			dp[i][j+shift] = (s2-s1+mod)%mod;
			pre[!now][j+shift] = (pre[!now][j+shift-1]+dp[i][j+shift])%mod;
		}
		now = !now;
	}
	
	/*for(int i=-k; i<=k; i++){
		printf("dp[%d][%d] = %I64d\n", t, i, dp[t][i+shift]);
	}*/
	
	pre[now][-t*k-1+shift] = 0;
	for(int i=-t*k; i<=t*k; i++) pre[now][i+shift] = (pre[now][i-1+shift]+dp[t][i+shift])%mod;

	ll ans=0;
	int tmp = a-b-1;
	for(int i=-t*k; i<=t*k; i++){
		/*for(int j=-k*t; j<=i+tmp; j++){
			ll sum = dp[t][i+shift]*dp[t][j+shift];
			ans = (ans+sum%mod)%mod;
		}*/
		ll s1 = i+tmp>t*k?pre[now][t*k+shift]:pre[now][i+tmp+shift];
		ll sum = (s1-pre[now][-k*t-1+shift]+mod)%mod;
		sum = dp[t][i+shift]*sum%mod;
		ans = (ans+sum)%mod; 
	} 
	printf("%I64d\n", ans);
	return 0;
}