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;
long long sol_value;
vector<long long> sol;
void chkans(vector<long long> vec) {
  long long v = 0;
  for (auto x : vec) {
    v ^= x;
  }
  if (v <= sol_value) {
    if (v == sol_value && vec[0] != sol[0] && vec[1] != sol[0] &&
        (int)(vec.size() + sol.size()) <= k) {
      sol_value = 0;
      sol.push_back(vec[0]);
      sol.push_back(vec[1]);
    } else if (v < sol_value) {
      sol_value = v;
      sol = vec;
    }
  }
}
int main() {
  cin >> l >> r >> k;
  int len = min(r - l + 1, 10ll);
  if (k == 1) {
    sol_value = l;
    sol = vector<long long>{l};
  } else {
    sol_value = l;
    sol = vector<long long>{l};
    for (int i = 0; i < len - 1; i++) {
      chkans(vector<long long>{l + i, l + i + 1});
    }
    if (k >= 3) {
      long long x = l;
      long long high = 0;
      for (int i = 50; i >= 0; i--) {
        if (x & (1ll << i)) {
          high = (1ll << i);
          break;
        }
      }
      long long a = l;
      long long c = high * 3;
      long long b = a ^ c;
      if (c <= r) {
        sol_value = a ^ b ^ c;
        sol = vector<long long>{a, b, c};
      }
    }
  }
  cout << sol_value << endl;
  cout << sol.size() << endl;
  for (auto v : sol) cout << v << " ";
  cout << endl;
  return 0;
}