/*
n<=5000
­这样就不能用O(n)的转移了，而是要用O(1)
的转移。
­注意我们每次的转移都来自一个连续的区间，
而且我们是求和
­区间求和？
­前缀和！
­令sum[step][i]表示f[step][1~i]的和
­还是以B下侧为例
­ f[step][i]=sum[step-1][i-1]+sum[step-1][k]-sum[step-1][i] 

*/
#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int MAXN = 6000;const int MOD  = 1000000007;int f[MAXN][MAXN] = {{0}};int n, a, b, k;inline int getnum(){    char c; int ans = 0; bool flag = false;    while ((c = getchar()) == ‘ ‘ || c == ‘\r‘ || c == ‘\n‘);    if (c == ‘-‘) flag = true; else ans = c - ‘0‘;    while ((c = getchar()) >= ‘0‘ && c <= ‘9‘) ans = ans * 10 + c - ‘0‘;    return ans * (flag ? -1 : 1);}inline void add(int &a, int b){    ll tmp = (ll)a + b;    if (tmp >= MOD) a = (int)(tmp - MOD);    else if (tmp < 0) a = (int)(tmp + MOD);    else a = (int)tmp;}int main(){    freopen("lift.in", "r", stdin);    freopen("lift.out", "w", stdout);    n = getnum(); a = getnum(); b = getnum(); k = getnum();    if (fabs(a - b) - 1 == 0) { printf("0\n"); return 0; }    for (int i = a; i <= n; i++)        f[0][i] = 1;    for (int step = 1; step <= k; step++)    for (int i = 1; i <= n; i++)        if (i == b) f[step][i] = f[step][i - 1];        else        {            f[step][i] = f[step][i - 1];            if (i < b)            {                int x = (b + i) / 2;                if ((b + i) % 2 == 0) x--;                add(f[step][i], f[step - 1][x]);                add(f[step][i], -f[step - 1][i]);                add(f[step][i], f[step - 1][i - 1]);            }            else            {                int x = (b + i) / 2;                add(f[step][i], f[step - 1][n]);                add(f[step][i], -f[step - 1][x]);                add(f[step][i], -f[step - 1][i]);                add(f[step][i], f[step - 1][i - 1]);            }        }    printf("%d\n", f[k][n]);}