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;
void Qmin(vector<long long> &a, const vector<long long> b) {
  if (accumulate(b.begin(), b.end(), 0LL, [&](long long x, long long y) {
        return x ^ y;
      }) < accumulate(a.begin(), a.end(), 0LL, [&](long long x, long long y) {
        return x ^ y;
      }))
    a = b;
}
int main() {
  long long l, r, k, x;
  cin >> l >> r >> k;
  vector<long long> ans;
  ans.push_back(l);
  if (k >= 2) {
    Qmin(ans, {l, l + 1});
    if (l + 2 <= r) Qmin(ans, {l + 1, l + 2});
  }
  if (k >= 3) {
    x = 1LL << (64 - __builtin_clzll(l));
    if (x + x / 2 <= r) ans = {x - 1, x + x / 2, x + x / 2 - 1};
  }
  if (k >= 4) {
    Qmin(ans, {l, l + 1, l + 2, l + 3});
    if (l + 4 <= r) Qmin(ans, {l + 1, l + 2, l + 3, l + 4});
  }
  cout << accumulate(ans.begin(), ans.end(), 0LL,
                     [&](long long x, long long y) { return x ^ y; })
       << '\n'
       << ans.size() << '\n';
  for (int i = 0; i < ans.size(); ++i) cout << ans[i] << ' ';
}