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.io.*;
import java.math.*;
import java.util.*;

public class Main{
    FastScanner in;
    PrintWriter out;

    public void solve()
    {
        long l = in.nextLong();
        long r = in.nextLong();
        int k = in.nextInt();

        ArrayList<Long> ans = new ArrayList<>();
        if (r-l+1 <= 10) {
            long d = r - l + 1;
            long res = r, mask = 1;
            for (int i = 1; i < (1 << d); i++) {
                long tmp = -1, cnt = 0;
                for (int j = 0; j < d; j++)
                if ((i >> j) % 2 == 1) {
                    if (tmp == -1)
                        tmp = l + j;
                    else
                        tmp ^= l + j;
                    cnt++;
                }
                if (cnt > k) continue;

                if (tmp < res) {
                    res = tmp;
                    mask = i;
                }
            }

            for (int i = 0; i < d; i++)
            if ((mask>>i)%2==1) {
                ans.add(l+i);
            }

        }
        else {
            if (k == 1)
                ans.add(l);
            else if(k == 2) {
                if (l % 2 == 1) l++;
                ans.add(l);
                ans.add(l+1);
            }
            else if(k >= 4) {
                if (l % 2 == 1) l++;
                ans.add(l);
                ans.add(l+1);
                ans.add(l+2);
                ans.add(l+3);
            }
            else {
                long x = 1;
                while (x < l) {
                    x *= 2;
                    x++;
                }

                if (x + x/2 + 2 > r) {
                    if (l % 2 == 1) l++;
                    ans.add(l);
                    ans.add(l+1);
                }
                else {
                    ans.add(x);
                    ans.add(x + x/2 + 1);
                    ans.add(x + x/2 + 2);
                }

            }
        }

        Long xor = ans.get(0);
        for (int i = 0; i < ans.size() - 1; i++) {
            xor ^= ans.get(i+1);
        }
        out.println(xor);
        out.println(ans.size());
        for (long x : ans)
            out.print(x + " ");
    }

    public void run()
    {
        in = new FastScanner();
        out = new PrintWriter(System.out);

        solve();

        out.close();
    }


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

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

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

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