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.util.Arrays;
import java.util.Scanner;

public class MainD {
	MyScanner sc = new MyScanner();
	Scanner sc2 = new Scanner(System.in);
	long start = System.currentTimeMillis();
	long fin = System.currentTimeMillis();
	final int MOD = 1000000007;
	int[] dx = { 1, 0, 0, -1 };
	int[] dy = { 0, 1, -1, 0 };

	void run() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		long x = sc.nextLong();
		long[] a = sc.nextLongArray(n);
		int[] bitC = new int[63];
		for (int i = 0; i < n; i++) {
			long ai = a[i];
			for (int j = 0; ai > 0; j++, ai >>= 1) {
				if ((ai & 1) == 1)
					bitC[j]++;
			}
		}
		long ans = 0;
		for (int i = 0; i < n; i++) {
			long ob = a[i];
			for (int j = 0; ob > 0; j++, ob >>= 1) {
				if ((ob & 1) == 1)
					bitC[j]--;
			}

			long add = a[i];
			for (int j = 0; j < k; j++) add *= x;
			long tmp = add;
			for (int j = 0; tmp > 0; j++, tmp>>= 1) {
				if ((tmp & 1) == 1) bitC[j]++;
			}

			long res = 0;
			for (int j = 0; j < 63; j++) {
				if (bitC[j] > 0) {
					res += 1L << j;
				}
			}
			ans = Math.max(ans, res);
	
			for (int j = 0; add > 0; j++, add >>= 1) {
				if ((add & 1) == 1) bitC[j]--;
			}
			ob = a[i];
			for (int j = 0; ob > 0; j++, ob >>= 1) {
				if ((ob & 1) == 1)
					bitC[j]++;
			}
		}
		System.out.println(ans);
	}

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

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

	void debug2(int[][] array) {
		for (int i = 0; i < array.length; i++) {
			for (int j = 0; j < array[i].length; j++) {
				System.out.print(array[i][j]);
			}
			System.out.println();
		}
	}

	boolean inner(int h, int w, int limH, int limW) {
		return 0 <= h && h < limH && 0 <= w && w < limW;
	}

	void swap(int[] x, int a, int b) {
		int tmp = x[a];
		x[a] = x[b];
		x[b] = tmp;
	}

	// find minimum i (a[i] >= border)
	int lower_bound(int a[], int border) {
		int l = -1;
		int r = a.length;
		while (r - l > 1) {
			int mid = (l + r) / 2;
			if (border <= a[mid]) {
				r = mid;
			} else {
				l = mid;
			}
		}
		// r = l + 1
		return r;
	}

	boolean palindrome(String s) {
		for (int i = 0; i < s.length() / 2; i++) {
			if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
				return false;
			}
		}
		return true;
	}

	class MyScanner {
		int nextInt() {
			try {
				int c = System.in.read();
				while (c != '-' && (c < '0' || '9' < c))
					c = System.in.read();
				if (c == '-')
					return -nextInt();
				int res = 0;
				do {
					res *= 10;
					res += c - '0';
					c = System.in.read();
				} while ('0' <= c && c <= '9');
				return res;
			} catch (Exception e) {
				return -1;
			}
		}

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

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

		String next() {
			try {
				StringBuilder res = new StringBuilder("");
				int c = System.in.read();
				while (Character.isWhitespace(c))
					c = System.in.read();
				do {
					res.append((char) c);
				} while (!Character.isWhitespace(c = System.in.read()));
				return res.toString();
			} catch (Exception e) {
				return null;
			}
		}

		int[] nextIntArray(int n) {
			int[] in = new int[n];
			for (int i = 0; i < n; i++) {
				in[i] = nextInt();
			}
			return in;
		}

		int[][] nextInt2dArray(int n, int m) {
			int[][] in = new int[n][m];
			for (int i = 0; i < n; i++) {
				in[i] = nextIntArray(m);
			}
			return in;
		}

		double[] nextDoubleArray(int n) {
			double[] in = new double[n];
			for (int i = 0; i < n; i++) {
				in[i] = nextDouble();
			}
			return in;
		}

		long[] nextLongArray(int n) {
			long[] in = new long[n];
			for (int i = 0; i < n; i++) {
				in[i] = nextLong();
			}
			return in;
		}

		char[][] nextCharField(int n, int m) {
			char[][] in = new char[n][m];
			for (int i = 0; i < n; i++) {
				String s = sc.next();
				for (int j = 0; j < m; j++) {
					in[i][j] = s.charAt(j);
				}
			}
			return in;
		}
	}
}