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, i, a, b, c, ok;
void one() { cout << l << '\n' << 1 << '\n' << l << '\n'; }
void two() {
  if (l < (l ^ r))
    cout << l << '\n' << 1 << '\n' << l << '\n';
  else
    cout << (l ^ r) << '\n' << 2 << '\n' << l << " " << r << '\n';
}
void three() {
  if ((l ^ (l + 1)) < ((l + 1) ^ (l + 2)))
    cout << (l ^ (l + 1)) << '\n' << 2 << '\n' << l << " " << l + 1 << '\n';
  else
    cout << ((l + 1) ^ (l + 2)) << '\n'
         << 2 << '\n'
         << l + 1 << " " << l + 2 << '\n';
}
int main() {
  cin.sync_with_stdio(false);
  cin >> l >> r >> k;
  if (k == 1) {
    one();
    return 0;
  }
  if (k == 2) {
    if (l == r)
      one();
    else if (l + 1 == r)
      two();
    else
      three();
    return 0;
  }
  if (k == 3) {
    if (l == r)
      one();
    else if (l + 1 == r)
      two();
    else {
      ok = 0;
      for (i = 0; i <= 40; i++) {
        a = 3 * (1LL << i);
        b = 3 * (1LL << i) - 1;
        c = 2 * (1LL << i) - 1;
        if (a <= r && c >= l) {
          ok = 1;
          break;
        }
      }
      if (ok)
        cout << (a ^ b ^ c) << '\n'
             << 3 << '\n'
             << a << " " << b << " " << c << '\n';
      else
        three();
    }
    return 0;
  }
  if (k >= 4) {
    if (l == r)
      one();
    else if (l + 1 == r)
      two();
    else if (l + 2 == r)
      three();
    else if (l + 3 == r) {
      ok = 0;
      for (i = 0; i <= 40; i++) {
        a = 3 * (1LL << i);
        b = 3 * (1LL << i) - 1;
        c = 2 * (1LL << i) - 1;
        if (a <= r && c >= l) {
          ok = 1;
          break;
        }
      }
      if (ok)
        cout << (a ^ b ^ c) << '\n'
             << 3 << '\n'
             << a << " " << b << " " << c << '\n';
      else {
        if (l & 1)
          cout << 1 << '\n' << 2 << '\n' << l + 1 << " " << l + 2 << '\n';
        else
          cout << 0 << '\n'
               << 4 << '\n'
               << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << '\n';
      }
    } else {
      if (l & 1)
        cout << 0 << '\n'
             << 4 << '\n'
             << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4 << '\n';
      else
        cout << 0 << '\n'
             << 4 << '\n'
             << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << '\n';
    }
  }
  return 0;
}