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;
long long best = 1e12 + 1;
vector<long long> ans;
long long get_xor(vector<long long> arr) {
  long long ans = 0;
  for (int i = 0; i < arr.size(); ++i) ans ^= arr[i];
  return ans;
}
void check(vector<long long> arr) {
  long long val = get_xor(arr);
  if (val < best) {
    best = val;
    ans = arr;
  }
}
int main() {
  cin >> l >> r >> k;
  k = min(k, r - l + 1);
  check({l});
  if (k >= 2) {
    check({l, l + 1});
    if (l + 2 <= r) check({l + 1, l + 2});
  }
  if (k >= 3) {
    long long h = 1;
    while (2 * h <= l) h *= 2;
    if (h * 3 <= r) check({l, h * 3, l ^ (h * 3)});
  }
  if (k >= 4) {
    check({l, l + 1, l + 2, l + 3});
    if (l + 4 <= r) check({l + 1, l + 2, l + 3, l + 4});
  }
  cout << best << '\n' << ans.size() << '\n';
  for (int i = 0; i < ans.size(); ++i)
    cout << ans[i] << " \n"[i == ans.size() - 1];
};