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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long l, r;
  int k;
  while (cin >> l >> r >> k) {
    if (k >= 4) {
      if (l % 2 && r - l + 1 >= 5) {
        cout << 0 << "\n"
             << 4 << "\n"
             << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4 << "\n";
        continue;
      } else if (l % 2 == 0 && r - l + 1 >= 4) {
        cout << "0\n4\n"
             << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << "\n";
        continue;
      }
      k = 3;
    }
    if (k == 3) {
      long long c = 3LL, b = 2LL, a = 1LL;
      int flag = 0;
      while (1) {
        if (c <= r && a >= l) {
          flag = 1;
          break;
        }
        if (c > r) break;
        a = a << 1 | 1;
        b = b << 1 | 1;
        c = c << 1;
      }
      if (flag) {
        cout << "0\n3\n" << a << " " << b << " " << c << "\n";
        continue;
      }
      k = 2;
    }
    if (k == 2) {
      int flag = 0;
      for (long long i = l; i <= l + 1; i++) {
        if (i % 2LL == 0 && i + 1LL <= r) {
          flag = 1;
          cout << "1\n2\n" << i << " " << i + 1LL << "\n";
        }
      }
      if (!flag) {
        if ((l ^ r) < l)
          cout << (l ^ r) << "\n" << 2 << "\n" << l << " " << r << "\n";
        else
          cout << l << "\n1\n" << l << "\n";
      }
    }
    if (k == 1) {
      cout << l << "\n" << 1 << "\n" << l << "\n";
      ;
      continue;
    }
  }
  return 0;
}