Problem: On a chessboard with a width of n and a height of n, rows are numbered from bottom to top from 1 to n, columns are numbered from left to right from 1 to n. Therefore, for each cell of the chessboard, you can assign the coordinates (r,c), where r is the number of the row, and c is the number of the column.

The white king has been sitting in a cell with (1,1) coordinates for a thousand years, while the black king has been sitting in a cell with (n,n) coordinates. They would have sat like that further, but suddenly a beautiful coin fell on the cell with coordinates (x,y)...

Each of the monarchs wanted to get it, so they decided to arrange a race according to slightly changed chess rules:

As in chess, the white king makes the first move, the black king makes the second one, the white king makes the third one, and so on. However, in this problem, kings can stand in adjacent cells or even in the same cell at the same time.

The player who reaches the coin first will win, that is to say, the player who reaches the cell with the coordinates (x,y) first will win.

Let's recall that the king is such a chess piece that can move one cell in all directions, that is, if the king is in the (a,b) cell, then in one move he can move from (a,b) to the cells (a + 1,b), (a - 1,b), (a,b + 1), (a,b - 1), (a + 1,b - 1), (a + 1,b + 1), (a - 1,b - 1), or (a - 1,b + 1). Going outside of the field is prohibited.

Determine the color of the king, who will reach the cell with the coordinates (x,y) first, if the white king moves first.

Input

The first line contains a single integer n (2 ≤ n ≤ 10^{18}) — the length of the side of the chess field.

The second line contains two integers x and y (1 ≤ x,y ≤ n) — coordinates of the cell, where the coin fell.

Output

In a single line print the answer "White" (without quotes), if the white king will win, or "Black" (without quotes), if the black king will win.

You can print each letter in any case (upper or lower).

Examples

Input

4
2 3


Output

White

Input

5
3 5


Output

Black

Input

2
2 2


Output

Black

Note

An example of the race from the first sample where both the white king and the black king move optimally:

  1. The white king moves from the cell (1,1) into the cell (2,2).
  2. The black king moves form the cell (4,4) into the cell (3,3).
  3. The white king moves from the cell (2,2) into the cell (2,3). This is cell containing the coin, so the white king wins.

<image>

An example of the race from the second sample where both the white king and the black king move optimally:

  1. The white king moves from the cell (1,1) into the cell (2,2).
  2. The black king moves form the cell (5,5) into the cell (4,4).
  3. The white king moves from the cell (2,2) into the cell (3,3).
  4. The black king moves from the cell (4,4) into the cell (3,5). This is the cell, where the coin fell, so the black king wins.

<image>

In the third example, the coin fell in the starting cell of the black king, so the black king immediately wins.

<image>

Solution (in JAVA):

Solution:
//Author: Patel Rag
//Java version "1.8.0_211"
import java.util.*;
import java.io.*;

public class Main
{
  static class FastReader
  {
    BufferedReader br;
    StringTokenizer st;

    public FastReader(){ br = new BufferedReader(new InputStreamReader(System.in)); }

    String next()
    {
      while (st == null || !st.hasMoreElements())
      {
        try
        {
          st = new StringTokenizer(br.readLine());
        }
        catch (IOException  e)
        {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() { return Integer.parseInt(next()); }

    long nextLong(){ return Long.parseLong(next()); }

    double nextDouble(){ return Double.parseDouble(next()); }

    float nextFloat() { return Float.parseFloat(next()); }

    boolean nextBoolean() { return Boolean.parseBoolean(next()); }

    String nextLine()
    {
      String str = "";
      try
      {
        str = br.readLine();
      }
      catch (IOException e)
      {
        e.printStackTrace();
      }
      return str;
    }
  }
  static long modExp(long x, long n, long mod)    //binary Modular exponentiation
  {
    long result = 1;
    while(n > 0)
    {
        if(n % 2 == 1)
            result = (result%mod * x%mod)%mod;
        x = (x%mod * x%mod)%mod;
        n=n/2;
    }
    return result;
  }
  static long gcd(long a, long b)
  {
    if(a==0) return b;
    return gcd(b%a,a);
  }
  public static void main(String[] args)
  throws IOException
  {
    FastReader fr = new FastReader();
    long n = fr.nextLong();
    long x = fr.nextLong();
    long y = fr.nextLong();
    long w = Long.min(x,y) - 1 + (x - Long.min(x,y)) + (y - Long.min(x,y));
    long b = n - Long.max(x,y) + (Long.max(x,y) - x) + (Long.max(x,y) - y);
    if(w <= b) System.out.println("White");
    else System.out.println("Black");

  }
}
class Pair<U, V>  // Pair class
{
    public final U first;       // first field of a Pair
    public final V second;      // second field of a Pair

    private Pair(U first, V second)
    {
      this.first = first;
      this.second = second;
    }

    @Override
    public boolean equals(Object o)
    {
      if (this == o) return true;

      if (o == null || getClass() != o.getClass()) return false;

      Pair<?, ?> pair = (Pair<?, ?>) o;

      if (!first.equals(pair.first)) return false;
      return second.equals(pair.second);
    }

    @Override
    public int hashCode()
    {
      return 31 * first.hashCode() + second.hashCode();
    }

    public static <U, V> Pair <U, V> of(U a, V b)
    {
      return new Pair<>(a, b);
    }
}
class myComp implements Comparator<Pair>
{
  public int compare(Pair a,Pair b)
  {
    if(a.first != b.first) return ((int)a.first - (int)b.first);
    if(a.second != b.second) return ((int)a.second - (int)b.second);
    return 0;
  }
}
class BIT       //Binary Indexed Tree aka Fenwick Tree
{
  public long[] m_array;

  public BIT(long[] dat)
  {
    m_array = new long[dat.length + 1];
    Arrays.fill(m_array,0);
    for(int i = 0; i < dat.length; i++)
    {
      m_array[i + 1] = dat[i];
    }
    for(int i = 1; i < m_array.length; i++)
    {
      int j = i + (i & -i);
      if(j < m_array.length)
      {
        m_array[j] = m_array[j] + m_array[i];
      }
    }
  }

  public final long prefix_query(int i)
  {
    long result = 0;
    for(++i; i > 0; i = i - (i & -i))
    {
      result = result + m_array[i];
    }
    return result;
  }

  public final long range_query(int fro, int to)
  {
    if(fro == 0)
    {
      return prefix_query(to);
    }
    else
    {
      return (prefix_query(to) - prefix_query(fro - 1));
    }
  }

  public void update(int i, long add)
  {
    for(++i; i < m_array.length; i = i + (i & -i))
    {
      m_array[i] = m_array[i] + add;
    }
  }
}