You are given n numbers a1, a2, ..., an. You can perform at most k operations. For each operation you can multiply one of the numbers by x. We want to make <image> as large as possible, where <image> denotes the bitwise OR. 

Find the maximum possible value of <image> after performing at most k operations optimally.

Input

The first line contains three integers n, k and x (1 ≤ n ≤ 200 000, 1 ≤ k ≤ 10, 2 ≤ x ≤ 8).

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

Output the maximum value of a bitwise OR of sequence elements after performing operations.

Examples

Input

3 1 2
1 1 1


Output

3


Input

4 2 3
1 2 4 8


Output

79

Note

For the first sample, any possible choice of doing one operation will result the same three numbers 1, 1, 2 so the result is <image>. 

For the second sample if we multiply 8 by 3 two times we'll get 72. In this case the numbers will become 1, 2, 4, 72 so the OR value will be 79 and is the largest possible result.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class D {
	
	static class Scanner{
		BufferedReader br=null;
		StringTokenizer tk=null;
		public Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		public String next() throws IOException{
			while(tk==null || !tk.hasMoreTokens())
				tk=new StringTokenizer(br.readLine());
			return tk.nextToken();
		}
		public int nextInt() throws NumberFormatException, IOException{
			return Integer.valueOf(next());
		}
		public long nextLong() throws NumberFormatException, IOException{
			return Long.valueOf(next());
		}
		public double nextDouble() throws NumberFormatException, IOException{
			return Double.valueOf(next());
		}
	}
	
	static int[] count;
	
	
	static long quitar(long todo, long c) {
		int idx = 0;
		while(c > 0) {
			if ((c & 1) == 1) {
				if (count[idx] == 1)
					todo = todo ^ (1<<idx);
			}
			idx++;
			c = c >> 1;
		}
		return todo;
	}
	
	public static void main(String args[]) throws NumberFormatException, IOException {
		Scanner sc = new Scanner();
		int N = sc.nextInt();
		int K = sc.nextInt();
		int x = sc.nextInt();
		long[] a = new long[N];
		long todo = 0;
		count = new int[64];
		for(int i = 0; i < N; i++) {
			a[i] = sc.nextLong();
			todo = todo | a[i];
			long copia = a[i];
			int idx = 0;
			while(copia > 0) {
				if ((copia & 1) == 1)
					count[idx]++;
				idx++;
				copia = copia >> 1;
			}
		}
		long ans = todo;
		long mult = 1;
		for(int i = 0; i < K; i++)
			mult *= x;
		for(int i = 0; i < N; i++) {
			long tmp = quitar(todo, a[i]);
			tmp = tmp | (a[i] * mult);
			ans = Math.max(ans, tmp);
		}
		System.out.println(ans);
	}

}
