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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
public class A {
 
	
	/* 10^(7) = 1s.
	 * ceilVal = (a+b-1) / b */
	
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	static class Pair implements Comparable<Pair> 
	{
		int first, second;
		public Pair(int aa, int bb)
		{
			first = aa; second = bb;
		}
		public int compareTo(Pair o)
		{
			if(this.second < o.second) return -1;
			if(this.second > o.second) return +1;
			return this.first - o.first;
		}
	}
	
	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank.
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	public static boolean isPrime(long n) 
	{
		if(n == 1)
		{
			return false;
		}
		for(int i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
}
	
	public static ArrayList<Integer> SieveList(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		
		ArrayList<Integer> l = new ArrayList<>();
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static int lcm(int a, int b)
	    {
	        return (a / gcd(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		public static long phi(long n)  //euler totient function
	    { 
	       long ans = n;
//	       for(long i = 2;i*i<=n;i++)
//	       {
//	    	   if(n%i == 0)
//	    	   {
//	    		   while(n%i == 0) n/=i;
//	    		   ans -= (ans/i);
//	    	   }
//	       }
//	       
//	       if(n > 1)
//	       {
//	    	   ans -= (ans/n);
//	       }
	       
	       for(long i = 2;i<=n;i++)
	       {
	    	   if(isPrime(i))
	    	   {
	    		   ans -= (ans/i);
	    	   }
	       }
	       
	       return ans;
	       
	    }
	
	public static int fastPow(int x, int n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	    	if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long fact(long n) {
	    	  long[] fac = new long[(int)(n) + 1];
		        fac[0] = 1;
		 
		        for (long i = 1; i <= n; i++)
		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
		        
		        return fac[(int)(n)];
		}
	    
	    public static long nCr(int n, int k) 
	    {
			long ans = 1;
			for(int i = 0;i<k;i++)
			{
				ans *= (n-i);
				ans /= (i+1);
			}
			return ans;
		}
 
		public static void Sort(int[] a) {
			List<Integer> l = new ArrayList<>();
			for (int i : a) l.add(i);
			Collections.sort(l);
			Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a)
		{
			List<Character> l = new ArrayList<>();
			for (char i : a) l.add(i);
			Collections.sort(l);
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		  
		   public static long subMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)-perfomMod(b));
		   }
		   
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		   
		   /*
		    * 
		    * >=         <=
		   0 1 2 3 4 5 6 7  
		   5 5 5 6 6 6 7 7
		   lower_bound for 6 at index 3 (>=)
		   upper_bound for 6 at index 6(To get six reduce by one) (<=)
		   */
		
		   public static int LowerBound(int a[], int x) 
		   {
				  int l=-1,r=a.length;
				  while(l+1<r) {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		
		   
			public static int UpperBound(int a[], int x) 
			{
			    int l=-1;
			    int r=a.length;
			    while(l+1<r) {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}
			
			
		
	public static void main(String[] args) throws IOException 
	{
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);

		long x = sc.nextLong();
		long one = sc.nextLong(), two = sc.nextLong();
		
		long fr = max(one-1,two-1);
		long sr = max(x - one,x - two);
		
		if(fr <= sr)
		{
			fout.println("White");
		}
		else
		{
			fout.println("Black");
		}
		
		fout.close();
	}
}