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.HashSet;
import java.util.Scanner;
import java.util.Set;

/**
 * Created by hama_du on 2014/08/24.
 */
public class ProblemD {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long l = in.nextLong();
        long r = in.nextLong();
        long k = in.nextInt();

        long min = l;
        Set<Long> bestValues = new HashSet<>();
        bestValues.add(l);

        // voe
        for (long w = 2 ; w <= Math.min(k, 4) ; w++) {
            long xor = 0;
            for (long x = l; x <= r ; x++) {
                if (Math.abs(x - l) >= 1000000) {
                    break;
                }
                xor ^= x;
                if (x - l >= w) {
                    xor ^= (x-w);
                }
                long reg = xor;
                if (min > reg) {
                    min = reg;
                    bestValues.clear();
                    for (long i = x-w+1 ; i <= x ; i++) {
                        bestValues.add(i);
                    }
                }
            }
        }

        // okay... try another solution.
        if (k >= 3 && Long.highestOneBit(l) < Long.highestOneBit(r)) {
            long hl = Long.highestOneBit(l);
            if(hl*3 <= r) {
                min = 0;
                bestValues.clear();
                bestValues.add(hl*2-1);
                bestValues.add(hl*3-1);
                bestValues.add(hl*3);
            }
        }


        System.out.println(min);
        System.out.println(bestValues.size());
        StringBuilder b = new StringBuilder();
        for (long i : bestValues) {
            b.append(' ');
            b.append(i);
        }
        System.out.println(b.substring(1));
    }




}