codeforces479E

Riding in a Lift

CodeForces - 479E

Imagine that you are in a building that has exactly n floors. You can move between the floors in a lift. Let's number the floors from bottom to top with integers from 1 to n. Now you're on the floor number a. You are very bored, so you want to take the lift. Floor number b has a secret lab, the entry is forbidden. However, you already are in the mood and decide to make k consecutive trips in the lift.

Let us suppose that at the moment you are on the floor number x (initially, you were on floor a). For another trip between floors you choose some floor with number y (y ≠ x) and the lift travels to this floor. As you cannot visit floor b with the secret lab, you decided that the distance from the current floor x to the chosen ymust be strictly less than the distance from the current floor x to floor b with the secret lab. Formally, it means that the following inequation must fulfill: |x - y| < |x - b|. After the lift successfully transports you to floor y, you write down number y in your notepad.

Your task is to find the number of distinct number sequences that you could have written in the notebook as the result of k trips in the lift. As the sought number of trips can be rather large, find the remainder after dividing the number by 1000000007 (10⁹ + 7).

Input

The first line of the input contains four space-separated integers n, a, b, k (2 ≤ n ≤ 5000, 1 ≤ k ≤ 5000, 1 ≤ a, b ≤ n, a ≠ b).

Output

Print a single integer — the remainder after dividing the sought number of sequences by 1000000007 (10⁹ + 7).

Examples

Input

5 2 4 1

Output

2

Input

5 2 4 2

Output

2

Input

5 3 4 1

Output

0

Note

Two sequences p₁, p₂, ..., p_k and q₁, q₂, ..., q_k are distinct, if there is such integer j (1 ≤ j ≤ k), that p_j ≠ q_j.

Notes to the samples:

1. In the first sample after the first trip you are either on floor 1, or on floor 3, because |1 - 2| < |2 - 4| and |3 - 2| < |2 - 4|.
2. In the second sample there are two possible sequences: (1, 2); (1, 3). You cannot choose floor 3 for the first trip because in this case no floor can be the floor for the second trip.
3. In the third sample there are no sought sequences, because you cannot choose the floor for the first trip.

sol：非常水的dp，直接转移是n³的，但是智障选手比方说我写了数据结构优化（树状数组），可以做到n²logn，但是还是TLE了，然后想了一会发现自己脑抽了，直接差分就是n²了

然后悲伤的T了两发

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline int read()
{
    int s=;
    bool f=;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<)+(s<<)+(ch^); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(int x)
{
    if(x<)
    {
        putchar('-'); x=-x;
    }
    if(x<)
    {
        putchar(x+'');    return;
    }
    write(x/);
    putchar((x%)+'');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=,Mod=;
int n,a,b,m;
int dp[N][N];
inline void Ad(int &x,int y)
{
    x+=y;
    x-=(x>=Mod)?Mod:;
    x+=(x<)?Mod:;
}
struct BIT
{
    int S[N];
    #define lowbit(x) ((x)&(-x))
    inline void Init()
    {
        memset(S,,sizeof S);
    }
    inline void Ins(int x,int Val)
    {
        for(;x<=n;x+=lowbit(x))
        {
            Ad(S[x],Val);
        }
    }
    inline void Updata(int l,int r,int Val)
    {
        Ins(l,Val); Ins(r+,-*Val);
    }
    inline int Que(int x)
    {
        int Sum=;
        for(;x>;x-=lowbit(x))
        {
            Ad(Sum,S[x]);
        }
        return Sum;
    }
}T;
int S[N];
int main()
{
    int i,j,k;
    R(n); R(a); R(b); R(m);
    dp[a][]=;
    for(i=;i<m;i++)
    {
//        T.Init();
        for(j=;j<=n;j++) if(dp[j][i])
        {
            int oo=abs(j-b)-;
//            if(max(1,j-oo)<=j-1) T.Updata(max(1,j-oo),j-1,dp[j][i]);
//            if(j+1<=min(n,j+oo)) T.Updata(j+1,min(n,j+oo),dp[j][i]);
            if(max(,j-oo)<=j-)
            {
                Ad(dp[max(,j-oo)][i+],dp[j][i]); Ad(dp[j-+][i+],-*dp[j][i]);
            }
            if(j+<=min(n,j+oo))
            {
                Ad(dp[j+][i+],dp[j][i]); Ad(dp[min(n,j+oo)+][i+],-*dp[j][i]);
            }
        }
        for(j=;j<=n;j++) Ad(dp[j][i+],dp[j-][i+]);
//        for(j=1;j<=n;j++) dp[j][i+1]=T.Que(j);
    }
    int ans=;
    for(i=;i<=n;i++) Ad(ans,dp[i][m]);
    Wl(ans);
    return ;
}
/*
Input
5 2 4 1
Output
2

Input
5 2 4 2
Output
2

Input
5 3 4 1
Output
0

input
2222 1206 1425 2222
output
402572650

Input
5000 2500 1 5000
Output
898026985
*/