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;
void print_ret(const std::vector<long long>& ret) {
  long long r = 0;
  for (typeof((ret).end()) it = (ret).begin(); it != (ret).end(); ++it)
    r ^= *it;
  cout << r << endl;
  cout << ret.size() << endl;
  for (typeof((ret).end()) it = (ret).begin(); it != (ret).end(); ++it)
    cout << *it << ' ';
  cout << endl;
}
int main(int argc, const char* argv[]) {
  long long l, r, k;
  std::vector<long long> ret;
  cin >> l >> r >> k;
  if (k == 1) {
    ret.push_back(l);
  } else if (k == 2) {
  k2:
    if (r - l >= 2) {
      ret.push_back(l + 1);
      if (l & 1)
        ret.push_back(l + 2);
      else
        ret.push_back(l);
    } else {
      ret.push_back(l);
      if ((l ^ (l + 1)) < l) ret.push_back(l + 1);
    }
  } else if (k == 3) {
    long long t = l;
    while (t & (t - 1)) t &= (t - 1);
    t |= (t << 1);
    if (t <= r) {
      ret.push_back(l);
      ret.push_back(t);
      ret.push_back(l ^ t);
    } else {
      goto k2;
    }
  } else {
    for (long long t = l; t <= r; t++) {
      if (!(t & 3)) {
        if (t + 3 <= r) {
          ret.push_back(t);
          ret.push_back(t + 1);
          ret.push_back(t + 2);
          ret.push_back(t + 3);
          print_ret(ret);
          return 0;
        }
        break;
      }
    }
    std::vector<long long> vals;
    for (long long t = l; t <= r; t++) vals.push_back(t);
    long long bestval = l;
    long long bestmask = 1;
    for (int mask = 1; mask < (1 << vals.size()); mask++) {
      long long cur = 0;
      for (int i = 0; i < vals.size(); i++) {
        if (mask & (1 << i)) cur ^= vals[i];
      }
      if (cur < bestval) {
        bestval = cur;
        bestmask = mask;
      }
    }
    for (int i = 0; i < vals.size(); i++) {
      if (bestmask & (1 << i)) ret.push_back(vals[i]);
    }
  }
  print_ret(ret);
  return 0;
}