Problem: Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties:

  * for all x <image> the following inequality holds l ≤ x ≤ r; 
  * 1 ≤ |S| ≤ k; 
  * lets denote the i-th element of the set S as si; value <image> must be as small as possible. 



Help Victor find the described set.

Input

The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)).

Output

Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order.

If there are multiple optimal sets, you can print any of them.

Examples

Input

8 15 3


Output

1
2
10 11


Input

8 30 7


Output

0
5
14 9 28 11 16

Note

Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation.

Solution (in JAVA):

Solution:
import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author xiaodao (www.shuizilong.com/house)
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {


    public void solve(int testNumber, Scanner in, PrintWriter out) {
        long l = in.nextLong(), r = in.nextLong(), k = in.nextLong();

        long rr = r;


        if (r > l + 20) r = l + 20;


        if (r <= l + 20) {

            long ans = r;

            long n = r - l + 1;
            for (int s = 1; s < 1 << n; ++s) {
                long xx = 0;
                int cnt = 0;
                for (int i = 0; i < n; ++i)
                    if ((s & (1 << i)) != 0) {
                        ++cnt;
                        xx ^= (l + i);
                    }
                if (cnt > k) continue;

                if (xx < ans) {
                    ans = xx;
                }
            }

            if (ans == 1 && k == 3){

                long key = 1; while (key <= l) key *= 2;

                long ll = key + (key >> 1);
                long lll = l ^ ll;
                if (ll <= rr && lll <= rr){
                    out.println(0);
                    out.println(3);
                    out.println(l+" "+ll +" " + lll);
                    return;
                }


            }


            out.println(ans);

            for (int s = 1; s < 1 << n; ++s) {
                long xx = 0;
                int cnt = 0;
                for (int i = 0; i < n; ++i)
                    if ((s & (1 << i)) != 0) {
                        ++cnt;
                        xx ^= (l + i);
                    }
                if (cnt > k) continue;

                if (xx == ans) {
                    out.println(cnt);
                    for (int i = 0; i < n; ++i)
                        if ((s & (1 << i)) != 0) {
                            out.print(l + i);
                            out.print(' ');
                        }
                    return;
                }
            }
        }

    }
}