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;
int main() {
  int i, j, k, bit;
  long long l, r, lc, ans, num, a, b, c;
  bool solve;
  vector<long long> u, v;
  cin >> l >> r >> k;
  ans = 1e15;
  if (r - l + 1 <= 10) {
    num = r - l + 1;
    for (i = 1; i < (1 << num); i++) {
      j = i;
      bit = 0;
      lc = 0;
      while (j) {
        if (j & 1) {
          u.push_back(l + bit);
          lc ^= (l + bit);
        }
        j >>= 1;
        bit++;
      }
      if (lc < ans && u.size() <= k) {
        v = u;
        ans = lc;
      }
      u.clear();
    }
  } else {
    if (k >= 4) {
      for (i = 0; i < 4; i++) {
        if ((l + i) % 4 == 0) break;
      }
      ans = 0;
      v.push_back(l + i);
      v.push_back(l + i + 1);
      v.push_back(l + i + 2);
      v.push_back(l + i + 3);
    } else if (k == 2) {
      if (l % 2 == 0) {
        v.push_back(l);
        v.push_back(l + 1);
      } else {
        v.push_back(l + 1);
        v.push_back(l + 2);
      }
      ans = 1;
    } else if (k == 3) {
      a = 1;
      b = 2;
      c = 3;
      solve = false;
      while (c <= r) {
        if (a >= l) {
          solve = true;
          v.push_back(a);
          v.push_back(b);
          v.push_back(c);
          ans = 0;
          break;
        }
        a = 2 * a + 1;
        b = 2 * b + 1;
        c = 2 * c;
      }
      if (!solve) {
        if (l % 2 == 0) {
          v.push_back(l);
          v.push_back(l + 1);
        } else {
          v.push_back(l + 1);
          v.push_back(l + 2);
        }
        ans = 1;
      }
    } else {
      ans = l;
      v.push_back(l);
    }
  }
  cout << ans << endl;
  cout << v.size() << endl;
  for (i = 0; i < v.size(); i++) cout << v[i] << " ";
}