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() {
  long long l, r, k;
  cin >> l >> r >> k;
  if (r - l <= 10) {
    vector<long long> v;
    for (long long i = l; i <= r; i++) {
      v.push_back(i);
    }
    int N = (int)(r - l + 1);
    long long ans = l;
    vector<long long> out;
    for (int m = 1; m < 1 << N; m++) {
      if (__builtin_popcount(m) <= k) {
        long long temp = 0;
        for (int k = 0; k < N; k++) {
          if ((m >> k) & 1) {
            temp ^= v[k];
          }
        }
        if (temp <= ans) {
          ans = temp;
          out.clear();
          for (int k = 0; k < N; k++) {
            if ((m >> k) & 1) {
              out.push_back(v[k]);
            }
          }
        }
      }
    }
    cout << ans << endl;
    cout << out.size() << endl;
    for (long long n : out) {
      cout << n << " ";
    }
  } else if (k == 1) {
    cout << l << "\n" << 1 << "\n" << l << endl;
  } else if (k == 2) {
    l = (l + 1) / 2 * 2;
    cout << 1 << "\n" << 2 << "\n" << l << " " << l + 1 << endl;
  } else if (k == 3) {
    long long z = 1, x = 3;
    while (x <= r) {
      if (z >= l) {
        cout << 0 << "\n" << 3 << "\n";
        cout << x << " " << z << " " << (x ^ z) << endl;
        return 0;
      }
      x *= 2;
      z *= 2;
      z++;
    }
    l = (l + 1) / 2 * 2;
    cout << 1 << "\n" << 2 << "\n" << l << " " << l + 1 << endl;
  } else {
    l = (l + 3) / 4 * 4;
    cout << 0 << "\n"
         << 4 << "\n"
         << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
  }
}