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, k;
  cin >> l >> r >> k;
  long long msb = 1;
  while (msb <= l) msb *= 2;
  msb /= 2;
  if (k == 1) {
    cout << l << "\n1\n" << l << '\n';
  } else if (k >= 4 && (r - l >= 4 || r % 2)) {
    cout << "0\n4\n";
    l = (l + 1) / 2 * 2;
    for (int i = 0; i <= 3; i++) cout << l + i << " \n"[i == 3];
  } else if (k >= 3 && msb * 3 <= r) {
    cout << "0\n3\n";
    cout << l << ' ' << 3 * msb << ' ' << msb + l << '\n';
  } else if (r - l >= 2 || r % 2) {
    cout << "1\n2\n";
    l = (l + 1) / 2 * 2;
    for (int i = 0; i <= 1; i++) cout << l + i << " \n"[i == 3];
  } else {
    assert(r - l == 1);
    long long alt = r ^ l;
    if (alt < l) {
      cout << alt << "\n2\n" << l << ' ' << r << '\n';
    } else {
      cout << l << "\n1\n" << l << '\n';
    }
  }
}