Problem: 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 y must 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, ..., pk and q1, q2, ..., qk are distinct, if there is such integer j (1 ≤ j ≤ k), that pj ≠ qj.

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. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
using namespace chrono;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int advceil(long long int num, long long int den) {
  return (num % den == 0 ? num / den : num / den + 1);
}
long long int lstbt(long long int val) {
  long long int msk = val & (val - 1);
  return log2(val ^ msk);
}
long long int modmul(long long int a, long long int b) {
  a %= 1000000007;
  b %= 1000000007;
  long long int res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % 1000000007;
    a = (a * 2) % 1000000007;
    b /= 2;
  }
  return res;
}
long long int modexpo(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = modmul(res, a);
    a = modmul(a, a);
    b /= 2;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return a == 0 ? b : gcd(b % a, a);
}
vector<long long int> CALCfactor(long long int n) {
  vector<long long int> factor(n + 2, 0);
  for (long long int i = 4; i <= n; i += 2) factor[i] = 2;
  for (long long int j = 3; j <= n; j += 2) {
    if (factor[j]) continue;
    for (long long int i = 2 * j; i <= n; i += j) factor[i] = j;
  }
  return factor;
}
vector<long long int> CALCprimeNUM(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> primeNUM;
  primeNUM.reserve(n + 2);
  for (long long int i = 2; i <= n; i++)
    if (!factor[i]) primeNUM.push_back(i);
  return primeNUM;
}
vector<long long int> CALCprimeFACTOR(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> ans;
  while (factor[n] != 0) {
    ans.push_back(factor[n]);
    n /= factor[n];
  }
  ans.push_back(n);
  return ans;
}
vector<long long int> unique(vector<long long int> x) {
  sort(x.begin(), x.end());
  set<long long int> s;
  vector<long long int> ans;
  ans.reserve(x.size());
  for (auto elem : x) s.insert(elem);
  for (auto elem : s) ans.push_back(elem);
  return ans;
}
pair<vector<long long int>, vector<long long int> > getFact(long long int n) {
  vector<long long int> fact(n + 1, 1), invfact(n + 1, 1);
  for (long long int i = 1; i <= n; i++)
    fact[i] = (i * (fact[i - 1])) % 1000000007;
  for (long long int i = 1; i <= n; i++)
    invfact[i] = (modexpo(i, 1000000007 - 2) * invfact[i - 1]) % 1000000007;
  return {fact, invfact};
}
long long int dp[5061][5061] = {};
void solve() {
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  for (long long int i = 1; i <= n; i++) dp[0][i] = 1;
  dp[0][b] = 0;
  for (long long int i = 1; i <= k; i++) {
    vector<long long int> pref(n + 2, 0);
    for (long long int j = 1; j <= n; j++)
      pref[j] = (pref[j - 1] + dp[i - 1][j]) % 1000000007;
    for (long long int j = 1; j <= n; j++) {
      long long int maxican = abs(j - b) - 1;
      if (maxican <= 0) continue;
      long long int L = max(1ll, j - maxican);
      long long int R = min(n, j + maxican);
      long long int curr = (pref[R] - pref[L - 1] + 1000000007) % 1000000007;
      dp[i][j] = (curr - dp[i - 1][j] + 1000000007) % 1000000007;
    }
  }
  cout << dp[k][a] % 1000000007 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start1 = high_resolution_clock::now();
  solve();
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
}