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;
template <class T>
T sqr(T x) {
  return x * x;
}
int main() {
  long long l, r, n, i, j, t, a1;
  vector<long long> ans;
  cin >> l >> r >> n;
  if (r - l + 1 >= 8) {
    if (n >= 4) {
      for (i = l; i <= r; i++)
        if (i % 2 == 0) {
          ans.push_back(i);
          ans.push_back(i + 1);
          if (ans.size() == 4) break;
        }
    } else if (n == 3) {
      int flag = 0;
      i = 1;
      while (i <= r) {
        i *= 2;
        if (i + i / 2 <= r && i - 1 >= l) {
          ans.push_back(i - 1);
          ans.push_back(i + i / 2 - 1);
          ans.push_back(i + i / 2);
          flag = 1;
          break;
        }
      }
      if (!flag)
        for (i = l; i <= r; i++)
          if (i % 2 == 0) {
            ans.push_back(i);
            ans.push_back(i + 1);
            break;
          }
    } else if (n == 2) {
      for (i = l; i <= r; i++)
        if (i % 2 == 0) {
          ans.push_back(i);
          ans.push_back(i + 1);
          break;
        }
    } else if (n == 1)
      ans.push_back(l);
  } else {
    for (i = 1; i < (1 << (r - l + 1)); i++) {
      if (__builtin_popcountll(i) > n) continue;
      vector<long long> now;
      long long t = 0;
      for (j = l; j <= r; j++)
        if ((1 << (j - l)) & i) {
          now.push_back(j);
          t ^= j;
        }
      if (ans.size() == 0 || t < a1) {
        ans = now;
        a1 = t;
      }
    }
  }
  a1 = 0;
  for (i = 0; i < ans.size(); i++) a1 ^= ans[i];
  cout << a1 << endl;
  printf("%d\n", ans.size());
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
}