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 JAVA):

Solution:
import java.util.*;

public class Main {
	static int mod = 1000000007;
  static int size = 200000;
	static long[] fac = new long[size];
	static long[] finv = new long[size];
	static long[] inv = new long[size];
	static int INF = Integer.MAX_VALUE;

 	public static void main(String[] args){
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int a = scanner.nextInt();
		int b = scanner.nextInt();
		int k = scanner.nextInt();
		//dp[i][j]:=i回目の移動後にj階にいる場合の数
		int[][] dp = new int[k+1][n+1];
		int[] sum = new int[5010];
		if(a < b){
			dp[0][a] = 1;
			for(int j = 1; j < b; j++){
				sum[j] = sum[j-1] + dp[0][j];
			}
			for(int i = 1; i <= k; i++){
				for(int j = 1; j < b; j++){
					dp[i][j] = (sum[j+(b-j-1)/2]-dp[i-1][j]+mod)%mod;
				}
				sum[0] = 0;
				for(int j = 1; j < b; j++){
					sum[j] = sum[j-1] + dp[i][j];
					sum[j] %= mod;
				}
			}
			System.out.println(sum[b-1]);
		}else{
			dp[0][a] = 1;
			for(int j = n; j >= b+1;j--){
				sum[j] += sum[j+1] + dp[0][j];
			}
			for(int i = 1; i <= k; i++){
				for(int j = b+1; j <= n; j++){
					dp[i][j] = (sum[j-(j-b-1)/2]-dp[i-1][j]+mod)%mod;
				}
				sum[0] = 0;
				for(int j = n; j >= b+1; j--){
					sum[j] = (sum[j+1] + dp[i][j]) % mod;
				}
			}
			System.out.println(sum[b+1]);
		}
	}
	public static boolean isPrime(int n){
		if(n == 1) return false;
		if(n == 2 || n == 3) return true;
		for(int i = 2; i <= Math.sqrt(n); i++){
			if(n % i == 0) return false;
		}
		return true;
	}
	// tar の方が数字が大きいかどうか
	static boolean compare(String tar, String src) {
		if (src == null) return true;
		if (src.length() == tar.length()) {
			int len = tar.length();
			for (int i = 0; i < len; i++) {
				if (src.charAt(i) > tar.charAt(i)) {
					return false;
				} else if (src.charAt(i) < tar.charAt(i)) {
					return true;
				}
			}
			return tar.compareTo(src) > 0 ? true : false;
		} else if (src.length() < tar.length()) {
			return true;
		} else if (src.length() > tar.length()) {
			return false;
		}
		return false;
	}
	public static class Edge{
		int to;
		Edge(int to){
			this.to = to;
		}
	}
	public static void swap(long a, long b){
		long tmp = 0;
		if(a > b){
			tmp = a;
			a = b;
			b = tmp;
		}
	}
  static class Pair implements Comparable<Pair>{
    int first, second;
    Pair(int a, int b){
        first = a;
        second = b;
    }
    @Override
    public boolean equals(Object o){
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return first == p.first && second == p.second;
    }
    @Override
    public int compareTo(Pair p){
        return first == p.first ? second - p.second : first - p.first; //firstで昇順にソート
        //return (first == p.first ? second - p.second : first - p.first) * -1; //firstで降順にソート
        //return second == p.second ? first - p.first : second - p.second;//secondで昇順にソート
        //return (second == p.second ? first - p.first : second - p.second)*-1;//secondで降順にソート
    }
  }

  //繰り返し二乗法
  public static long pow(long x, long n){
    long ans = 1;
    while(n > 0){
      if((n & 1) == 1){
        ans = ans * x;
        ans %= mod;
      }
      x = x * x % mod;
      n >>= 1;
    }
    return ans;
  }

	public static long div(long x, long y){
		return (x*pow(y, mod-2))%mod;
	}

  //fac, inv, finvテーブルの初期化、これ使う場合はinitComb()で初期化必要
	public static  void initComb(){
		fac[0] = finv[0] = inv[0] = fac[1] = finv[1] = inv[1] = 1;
		for (int i = 2; i < size; ++i) {
			fac[i] = fac[i - 1] * i % mod;
			inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
			finv[i] = finv[i - 1] * inv[i] % mod;
		}
	}

	//nCk % mod
	public static long comb(int n, int k){
		return fac[n] * finv[k] % mod * finv[n - k] % mod;
	}

	//n! % mod
	public static long fact(int n){
		return fac[n];
	}

	//(n!)^-1 with % mod
	public static long finv(int n){
		return finv[n];
	}

  static class UnionFind {
    int[] parent;
    public UnionFind(int size) {
      parent = new int[size];
      Arrays.fill(parent, -1);
    }
    public boolean unite(int x, int y) {
      x = root(x);
      y = root(y);
      if (x != y) {
        if (parent[y] < parent[x]) {
          int tmp = y;
          y = x;
          x = tmp;
        }
        parent[x] += parent[y];
        parent[y] = x;
        return true;
      }
      return false;
    }
    public boolean same(int x, int y) {
      return root(x) == root(y);
    }
    public int root(int x) {
      return parent[x] < 0 ? x : (parent[x] = root(parent[x]));
    }
    public int size(int x) {
      return -parent[root(x)];
    }
  }
	public static int upperBound(long[] array, long value) {
			 int low = 0;
			 int high = array.length;
			 int mid;
			 while( low < high ) {
					 mid = ((high - low) >>> 1) + low; // (high + low) / 2
					 if( array[mid] <= value ) {
							 low = mid + 1;
					 } else {
							 high = mid;
					 }
			 }
			 return low;
	 }
	 public static final int lowerBound(final int[] arr, final int value) {
    int low = 0;
    int high = arr.length;
    int mid;
    while (low < high) {
        mid = ((high - low) >>> 1) + low;    //(low + high) / 2 (オーバーフロー対策)
        if (arr[mid] < value) {
            low = mid + 1;
        } else {
            high = mid;
        }
    }
    return low;
}
  //n,mの最大公約数
  public static long gcd(long n, long m){
    if(m > n) return gcd(m,n);
    if(m == 0) return n;
    return gcd(m, n%m);
  }
	//3要素のソート
	private class Pair2 implements Comparable<Pair2> {
		String s;
		int p;
		int index;
		public Pair2(String s, int p, int index) {
				this.s = s;
				this.p = p;
				this.index = index;
		}

		public int compareTo(Pair2 other) {
				if (s.equals(other.s)) {
						return other.p - this.p;
				}
				return this.s.compareTo(other.s);
		}
	}
	//c -> intに変換
public static long c2i(char c){
	if('A' <= c && c <= 'Z'){
		return c - 'A';
	}else{
		return c - 'a' + 26;
	}
}
// int -> charに変換
public static char i2c(long i){
	if(0 <= i && i < 26){
			return (char)(i + 'A');
	}else{
			return (char)(i + 'a' - 26);
	}
}

}