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;
void f2() {}
int main() {
  cin >> l >> r >> k;
  if (k >= 4) {
    long long x = (l - 1) / 2 * 2 + 2;
    if (x >= l && x + 3 <= r) {
      cout << (x ^ (x + 1) ^ (x + 2) ^ (x + 3)) << endl;
      cout << 4 << endl;
      cout << x << " " << x + 1 << " " << x + 2 << " " << x + 3 << endl;
      return 0;
    }
  }
  if (k >= 3) {
    for (int i = 0; i < 100; i++) {
      long long x0 = 3ll << i;
      long long x1 = (3ll << i) - 1;
      long long x2 = (2ll << i) - 1;
      if (x0 > r) {
        break;
      }
      if (x2 < l) {
        continue;
      }
      cout << 0 << endl;
      cout << 3 << endl;
      cout << x2 << " " << x1 << " " << x0 << endl;
      return 0;
    }
  }
  if (l == 1) {
    cout << l << endl;
    cout << 1 << endl;
    cout << l << endl;
    return 0;
  }
  if (k >= 2) {
    long long x = (l - 1) / 2 * 2 + 2;
    if (x + 1 <= r) {
      cout << 1 << endl;
      cout << 2 << endl;
      cout << x << " " << (x + 1) << endl;
      return 0;
    }
  }
  if (k == 1 || l == r || (l ^ r) > l) {
    cout << l << endl;
    cout << 1 << endl;
    cout << l << endl;
    return 0;
  }
  cout << (l ^ r) << endl;
  cout << 2 << endl;
  cout << l << " " << r << endl;
  return 0;
}