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;
int main(void) {
  cin >> l >> r >> k;
  if (k == 1 || r - l + 1 == 1) {
    cout << l << "\n";
    cout << 1 << "\n";
    cout << l << "\n";
  } else if (k >= 4 && r - l + 1 >= 5) {
    long long i;
    for (i = l; i % 2 != 0; ++i)
      ;
    long long j;
    for (j = i + 1; j % 2 != 0; ++j)
      ;
    cout << 0 << "\n";
    cout << 4 << "\n";
    cout << i << " " << i + 1 << " " << j << " " << j + 1 << "\n";
  } else if (k >= 4 && r - l + 1 == 4) {
    long long i;
    for (i = l; i % 2 != 0; ++i)
      ;
    long long j;
    for (j = i + 1; j % 2 != 0; ++j)
      ;
    if (j + 1 > r) {
      if ((l ^ (l + 2) ^ (l + 3)) == 0) {
        cout << 0 << "\n";
        cout << 3 << "\n";
        cout << l << " " << l + 2 << " " << l + 3 << "\n";
      } else if ((i ^ (i + 1)) < l) {
        cout << 1 << "\n";
        cout << 2 << "\n";
        cout << i << ' ' << i + 1 << "\n";
      } else {
        cout << l << "\n";
        cout << 1 << "\n";
        cout << l << "\n";
      }
    } else {
      cout << 0 << "\n";
      cout << 4 << "\n";
      cout << i << ' ' << i + 1 << ' ' << j << ' ' << j + 1 << "\n";
    }
  } else if (k == 2 && r - l + 1 >= 3) {
    long long i;
    for (i = l; i % 2 != 0; ++i)
      ;
    cout << 1 << "\n";
    cout << 2 << "\n";
    cout << i << " " << i + 1 << "\n";
  } else if (r - l + 1 == 2) {
    if ((l ^ (l + 1)) < l) {
      cout << (l ^ (l + 1)) << "\n";
      cout << 2 << "\n";
      cout << l << " " << l + 1 << "\n";
    } else {
      cout << l << "\n";
      cout << 1 << "\n";
      cout << l << "\n";
    }
  } else if (k == 3) {
    if (2 * l > r || ((2 * l) ^ l) > r) {
      long long n1, n2;
      long long x = l;
      long long pot = 1;
      while (x) {
        x /= 2;
        pot *= 2;
      }
      n1 = pot;
      x = l;
      pot = 1;
      while (x) {
        if (x % 2 && n1 + pot <= r) {
          n1 += pot;
        }
        x /= 2;
        pot *= 2;
      }
      n2 = n1 ^ l;
      if (n1 >= l && n1 <= r && n2 >= l && n2 <= r) {
        cout << 0 << "\n";
        cout << 3 << "\n";
        cout << n1 << ' ' << l << ' ' << n2 << "\n";
      } else {
        long long i;
        for (i = l; i % 2 != 0; ++i)
          ;
        cout << 1 << "\n";
        cout << 2 << "\n";
        cout << i << " " << i + 1 << "\n";
      }
    } else {
      cout << 0 << "\n";
      cout << 3 << "\n";
      cout << l << ' ' << 2 * l << ' ' << ((2 * l) ^ l) << "\n";
    }
  }
  return 0;
}