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;
pair<unsigned long long, unsigned long long> get(unsigned long long n) {
  int tz = 63 - __builtin_clzll(n);
  unsigned long long a = n, b = 0;
  a ^= (1LL << (tz + 1));
  a ^= (1LL << (tz));
  b ^= (1LL << (tz + 1));
  b ^= (1LL << (tz));
  return {a, b};
}
int main() {
  unsigned long long l, r, k;
  cin >> l >> r >> k;
  unsigned long long tm = r - l + 1;
  vector<unsigned long long> ans;
  if (k == 1) {
    ans = {l};
  } else if (k == 2) {
    if (l % 2 == 0)
      ans = {l, l + 1};
    else if (tm > 2)
      ans = {l + 1, l + 2};
    else if (l < (l ^ r))
      ans = {l};
    else
      ans = {l, r};
  } else if (k == 3) {
    auto ns = get(l);
    if (ns.first <= r && ns.second <= r)
      ans = {l, ns.first, ns.second};
    else if (l % 2 == 0)
      ans = {l, l + 1};
    else
      ans = {l + 1, l + 2};
  } else if (k == 4) {
    auto ns = get(l);
    if (ns.first <= r && ns.second <= r)
      ans = {l, ns.first, ns.second};
    else if (l % 2 == 0)
      ans = {l, l + 1, l + 2, l + 3};
    else if (tm > 4)
      ans = {l + 1, l + 2, l + 3, l + 4};
    else
      ans = {l + 1, l + 2};
  } else {
    if (l % 2 == 0)
      ans = {l, l + 1, l + 2, l + 3};
    else
      ans = {l + 1, l + 2, l + 3, l + 4};
  }
  unsigned long long xx = 0;
  for (auto &i : ans) xx ^= i;
  cout << xx << endl;
  cout << ans.size() << endl;
  for (auto &i : ans) cout << i << ' ';
  cout << endl;
  return 0;
}