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;
const double eps(1e-8);
long long l, r, k;
int main() {
  cin >> l >> r >> k;
  vector<long long> ans;
  vector<long long> vec;
  if (k >= 5) {
    bool flag = false;
    for (long long i = l; i <= r; i++) {
      ans.clear();
      long long tmp = 0;
      for (long long j = i; j <= i + 4; j++) {
        ans.push_back(j);
        tmp ^= j;
        if (tmp == 0) {
          flag = true;
          break;
        }
      }
      if (flag) break;
    }
    cout << 0 << endl;
    cout << ans.size() << endl;
    for (int i = 0; i < (int)ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  } else {
    long long Min = (long long)1e15;
    for (long long i = l; i <= min(l + 100, r); i++) {
      vec.clear();
      long long tmp = 0;
      for (long long j = i; j <= min(i + k - 1, r); j++) {
        vec.push_back(j);
        tmp ^= j;
        if (tmp < Min) {
          Min = tmp;
          ans = vec;
        }
      }
      if (k >= 3) {
        vec.clear();
        long long x = 1;
        while (x <= i) x <<= 1;
        if (x + (x >> 1) <= r) {
          x += (x >> 1);
          Min = 0;
          vec.push_back(i);
          vec.push_back(x);
          vec.push_back(x ^ i);
          ans = vec;
          break;
        }
      }
    }
    cout << Min << endl;
    cout << ans.size() << endl;
    for (int i = 0; i < (int)ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  return 0;
}