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

D. Memory and Scores

Memory and his friend Lexa is 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 the current scores. The game has exactly T turns. Memory and Lexa, however, is not good at the this game, so they both always get a random integer at their turn.

Memory wonders how many possible games exist such, he ends with a strictly higher score than Lexa. Both games is considered to being different if in at least one turn at least one player gets different score. There is (2k + 1)2t games in total. Since the answer can be very large and you should print it modulo 9 + 7. Please solve the problem for Memory.

Input

The first and only line of input contains the four integers a, b, K, and t< /c7> (1≤ a, b ≤100, 1≤ k ≤1000, 1≤ t ≤100)-the amount Memo Ry and Lexa start with, the number K, and the number of turns respectively.

Output

Print the number of possible games satisfying the conditions modulo 1 007 (9 + 7) in one line.

Examples input

1 2 2 1

Output

6

Note

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 is3 + 2 + 1 = 6 possible games in which Memory wins.

Test instructions

A, b Two people play T-wheel game

No one can get any score from [-k,k] per round of games

AB starting points are a, B, respectively

Ask you the final a score is more rigorous than the number of programs in B

Exercises

Set DP[I][J] The number of schemes to obtain fractional J for the first round

This can be done with scrolling arrays and prefixes and optimizations

Finally enumerate a person's score to get the answer

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<queue>#include<Set>using namespacestd;#pragmaComment (linker, "/stack:102400000,102400000")#defineLS i<<1#defineRS ls | 1#defineMid ((LL+RR) >>1)#definePII pair<int,int>#defineMP Make_pairtypedefLong LongLL;Const Long LongINF =1e18;Const DoublePi = ACOs (-1.0);Const intN = 5e5+Ten, M = 1e2+ One, mod = 1e9+7, INF =2e9;inta,b,k,t; LL Sum[n], dp[2][n];intMain () {scanf ("%d%d%d%d",&a,&b,&k,&t); intnow =1; intLast = now ^1; dp[last][0] =1;  for(inti =0; I <=2* k * t; ++i) Sum[i] =1;  for(inti =1; I <= t; ++i) { for(intj =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(intj =1; J <=4* k * t; ++j) Sum[j]= ((sum[j-1] + dp[now][j])% mod + MoD)%MoD; now^=1; } LL ans=0;  for(inti =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;}

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