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 int l, r;
vector<long long int> ans;
int k;
long long int solve() {
  long long int res = l;
  ans.push_back(l);
  int diff = r - l + 1;
  for (int i = 1; i < 1 << diff; i++) {
    if (__builtin_popcount(i) <= k) {
      long long int _xor = 0;
      vector<long long int> tmp;
      for (int j = 0; j < diff; j++)
        if ((i >> j) & 1) {
          _xor ^= l + j;
          tmp.push_back(l + j);
        }
      if (_xor < res) {
        res = _xor;
        ans = tmp;
      }
    }
  }
  return res;
}
int main() {
  cin >> l >> r >> k;
  k = min(k, 4);
  if (k == 1 || r == l) {
    cout << l << "\n1\n" << l << endl;
    return 0;
  }
  if (r - l <= 6) {
    long long int val = solve();
    cout << val << "\n";
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << endl;
    return 0;
  }
  if (k == 3) {
    long long int x = 3;
    while (x <= r) {
      long long int y = x - 1;
      long long int z = (x / 3) * 2 - 1;
      if (z >= l && x <= r) {
        cout << "0\n3\n" << x << " " << y << " " << z << " " << endl;
        return 0;
      }
      x *= 2;
    }
    k = 2;
  }
  if (k == 2) {
    cout << "1\n2\n" << (l + 1) / 2 * 2 << " " << (l + 1) / 2 * 2 + 1 << "\n";
    return 0;
  }
  if (k == 4) {
    printf("0\n4\n");
    for (int i = 0; i < 4; i++) cout << (l + 3) / 4 * 4 + i << " ";
    cout << "\n";
  }
  return 0;
}