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 _min(int pos) {
  long long res = 1;
  for (pos--; pos >= 0; pos--) {
    res = res << 1;
  }
  return res;
}
long long aa(int pos, int pos2) {
  long long res = 1;
  for (int i = pos - 1; i >= 0; i--) {
    res <<= 1;
    if (i == pos2) res |= 1;
  }
  return res;
}
long long _max(int pos) {
  long long res = 1LL << (pos + 1);
  res--;
  return res;
}
bool solve() {
  int pos2;
  for (int pos = 39; pos >= 0; pos--) {
    if (_min(pos) <= r) {
      for (pos2 = pos - 1; pos2 >= 0; pos2--) {
        if (_max(pos2) >= l && aa(pos, pos2) <= r) {
          cout << "0" << endl;
          cout << "3" << endl;
          cout << aa(pos, pos2) << " " << _min(pos) + _max(pos2 - 1) << " "
               << _max(pos2) << endl;
          return true;
        }
      }
    }
  }
  return false;
}
int main() {
  while (cin >> l >> r >> k) {
    if (k == 1) {
      cout << l << endl;
      cout << "1" << endl;
      cout << l << endl;
      continue;
    } else if (k == 2) {
      if (r - l + 1 == 2) {
        if (l & 1) {
          long long ab = l ^ r;
          if (l > ab) {
            cout << ab << endl;
            cout << "2" << endl;
            cout << l << " " << r << endl;
          } else {
            cout << l << endl;
            cout << "1" << endl;
            cout << l << endl;
          }
        } else {
          cout << "1" << endl;
          cout << "2" << endl;
          cout << l << " " << l + 1 << endl;
        }
      } else {
        if (l & 1) l++;
        cout << "1" << endl;
        cout << "2" << endl;
        cout << l << " " << l + 1 << endl;
      }
      continue;
    } else if (k >= 4) {
      if (r - l + 1 == 4) {
        if (l & 1) {
          long long ab = l ^ (l + 1) ^ (l + 2) ^ (l + 3);
          if (solve())
            ;
          else {
            if (ab > 1) {
              cout << "1" << endl;
              cout << "2" << endl;
              cout << l + 1 << " " << l + 2 << endl;
            } else {
              cout << ab << endl;
              cout << "4" << endl;
              cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
            }
          }
        } else {
          cout << "0" << endl;
          cout << "4" << endl;
          cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
        }
      } else {
        if (l & 1) l++;
        cout << "0" << endl;
        cout << "4" << endl;
        cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
      }
      continue;
    } else if (k == 3) {
      if (!solve()) {
        if (l & 1) l++;
        cout << "1" << endl;
        cout << "2" << endl;
        cout << l << " " << l + 1 << endl;
      }
      continue;
    }
  }
  return 0;
}