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. 
import java.util.*;
import java.io.*;

public class Main implements Runnable {
	static int MOD = (int)(1e+9) + 7;
        
	public void solve() throws IOException {
		int N = nextInt(); int start = nextInt() - 1; int stop = nextInt() - 1; int K = nextInt();
                
                if(start > stop){//rotate the graph 180 degreee
                    start = N - 1 - start;
                    stop = N - 1 - stop;
                }
                
                long[] dp = new long[stop]; // no way we will be going go above STOP
                dp[start] = 1;
                
                long[] dp1 = new long[stop+1];
                for(int k = 0; k < K; k++){
                    
                    for(int i = 0; i < stop; i++){
                        long value = dp[i];
                        if(value == 0) continue;
                        
                        //the lowest we can reach AKA {i - Math.abs(i - stop} ;;; don't worry about other points,. because we would do cummulation later, hence this value will be added to other (possible destinations) at that instant
                        int j = 2 * i - (stop - 1);
                        if(j < 0) j = 0;
                        dp1[j] += value;
                        if(dp1[j] >= MOD) dp1[j] -= MOD;
                    }
                    
                    //now do cummulation thus we will be reaching all the way up
                    for(int i = 1; i < stop; i++){
                        dp1[i] += dp1[i-1];
                        if(dp1[i] >= MOD) dp1[i] -= MOD;
                    }
                    
                    //NOTE: we start at X and end at Y; given in problem X != Y, {there is a hole between cummulatives, we have already added that, so have to remove that}
                    // -dp[x] equals {MOD - dp[x]}
                    for(int i = 0; i < stop; i++){
                        dp1[i] = (dp1[i] + (MOD - dp[i])); //take out that value
                        if(dp1[i] >= MOD) dp1[i] -= MOD;   
                        
                        dp[i] = dp1[i]; 
                        dp1[i] = 0;
                    }
                }
                
                
                ///we might end at any of the point b/w 0.....stop
                long answer = 0;
                for(int i = 0; i < stop ; i++) answer += dp[i];
                
                out.println(answer % MOD);
                
	}

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	//-----------------------------------------------------------
	public static void main(String[] args) {
		new Main().run();
	}

        public void debug(Object... arr){
            System.out.println(Arrays.deepToString(arr));
        }

        public void print1Int(int[] a){
                for(int i = 0; i < a.length; i++)
                        System.out.print(a[i] + " ");
                System.out.println();
        }
        
        public void print2Int(int[][] a){
                for(int i = 0; i < a.length; i++){
                        for(int j = 0; j < a[0].length; j++){
                                System.out.print(a[i][j] + " ");
                        }
                        System.out.println();
                }
        }
        
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
                        out = new PrintWriter(System.out);
			tok = null;
			solve();
			in.close();
                        out.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

	public String nextToken() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
        PrintWriter out;
	BufferedReader in;
	StringTokenizer tok;
}