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;
long long l, r, k;
int main() {
  cin >> l >> r >> k;
  if (k >= 4) {
    if (l & 1 && l + 4 <= r) l += 1;
    if (!(l & 1) && l + 3 <= r) {
      puts("0\n4");
      for (int i = 0; i < 4; i++) cout << l + i << " ";
      cout << endl;
      return 0;
    }
  }
  if (k >= 3) {
    long long bit = 0, now = l;
    while (now) {
      now >>= 1;
      bit += 1;
    }
    now = 3;
    while (--bit) now <<= 1;
    if (now <= r) {
      puts("0\n3");
      cout << now << " " << (now ^ l) << " " << l << endl;
      return 0;
    }
  }
  if (k >= 2) {
    if (l % 2 == 0) {
      puts("1\n2");
      cout << l << " " << l + 1 << endl;
      return 0;
    }
    if (l % 2 == 1 && (l + 2 <= r)) {
      puts("1\n2");
      cout << l + 1 << " " << l + 2 << endl;
      return 0;
    }
    if ((l ^ r) < l) {
      cout << (l ^ r) << endl << 2 << endl << l << " " << r << endl;
      return 0;
    }
  }
  cout << l << endl << 1 << endl << l << endl;
  return 0;
}