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;
vector<long long> ans;
long long fs;
int main() {
  ios::sync_with_stdio(false);
  cin >> l >> r >> k;
  long long x = l / 4;
  long long rx = l % 4;
  long long y = r / 4;
  long long ry = r % 4;
  long long xx = x;
  long long yy = y;
  if (rx != 0) xx++;
  if (k == 1)
    ans.push_back(l);
  else if (k == 2) {
    if (l == r) {
      ans.push_back(l);
    } else if (r - l + 1 == 2) {
      if ((l ^ r) < l) {
        ans.push_back(l);
        ans.push_back(r);
      } else
        ans.push_back(l);
    } else {
      if (r % 4 == 0 || r % 4 == 2) {
        ans.push_back(r - 1);
        ans.push_back(r - 2);
      } else {
        ans.push_back(r);
        ans.push_back(r - 1);
      }
    }
  } else if (k == 3) {
    if (l == r) {
      ans.push_back(l);
    } else if (r - l + 1 == 2) {
      if ((l ^ r) < l) {
        ans.push_back(l);
        ans.push_back(r);
      } else
        ans.push_back(l);
    } else {
      bool sucess = false;
      for (long long i = l; i < r; i <<= 1) {
        long long x = i ^ r;
        if (x >= l && x <= r && x != i && x != r) {
          ans.push_back(i);
          ans.push_back(x);
          ans.push_back(r);
          sucess = true;
          break;
        }
      }
      if (!sucess) {
        if (r % 4 == 0 || r % 4 == 2) {
          ans.push_back(r - 1);
          ans.push_back(r - 2);
        } else {
          ans.push_back(r);
          ans.push_back(r - 1);
        }
      }
    }
  } else if (k >= 4) {
    if (l == r) {
      ans.push_back(l);
    } else if (r - l + 1 == 2) {
      if ((l ^ r) < l) {
        ans.push_back(l);
        ans.push_back(r);
      } else
        ans.push_back(l);
    } else {
      if (xx < yy || (xx == yy && ry == 3)) {
        ans.push_back(4 * xx);
        ans.push_back(4 * xx + 1);
        ans.push_back(4 * xx + 2);
        ans.push_back(4 * xx + 3);
      } else {
        bool sucess = false;
        for (long long i = l; i < r; i <<= 1) {
          long long x = i ^ r;
          if (x >= l && x <= r && x != i && x != r) {
            ans.push_back(i);
            ans.push_back(x);
            ans.push_back(r);
            sucess = true;
            break;
          }
        }
        if (!sucess) {
          if (rx <= 2 && ry >= 1) {
            ans.push_back(4 * xx);
            ans.push_back(4 * xx + 1);
            ans.push_back(4 * xx - 1);
            ans.push_back(4 * xx - 2);
          } else {
            if (r % 4 == 0 || r % 4 == 2) {
              ans.push_back(r - 1);
              ans.push_back(r - 2);
            } else {
              ans.push_back(r);
              ans.push_back(r - 1);
            }
          }
        }
      }
    }
  }
  for (auto &it : ans) fs ^= it;
  cout << fs << endl;
  cout << ans.size() << endl;
  for (auto &it : ans) cout << it << ' ';
  return 0;
}