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;
int main() {
  cin >> l >> r >> k;
  long long mn = l;
  vector<long long> res;
  res.push_back(l);
  if (l == r or k == 1) {
    cout << l << "\n" << 1 << "\n" << l << "\n";
    return 0;
  }
  if (k >= 2) {
    if (l + 1 == r) {
      if (l < (l ^ r)) {
        res.clear();
        mn = l;
        res.push_back(l);
      } else {
        res.clear();
        mn = l ^ r;
        res.push_back(l);
        res.push_back(r);
      }
    } else {
      long long v[] = {l, (l ^ (l + 1)), (l ^ (l + 2)), ((l + 1) ^ (l + 2))};
      sort(v, v + 4);
      if (v[0] < mn) {
        mn = v[0];
        res.clear();
        if (v[0] == l) {
          res.push_back(l);
        } else if (v[0] == (l ^ (l + 1))) {
          res.push_back(l);
          res.push_back(l + 1);
        } else if (v[0] == (l ^ (l + 2))) {
          res.push_back(l);
          res.push_back(l + 2);
        } else if (v[0] == ((l + 1) ^ (l + 2))) {
          res.push_back(l + 1);
          res.push_back(l + 2);
        }
      }
    }
  }
  if (k >= 3) {
    long long cnt = 0, tmp = l;
    while (tmp > 1) {
      cnt++;
      tmp >>= 1;
    }
    while (cnt >= 0) {
      cnt--;
      tmp <<= 1;
    }
    bool ok = true;
    long long a = tmp, b = tmp;
    for (long long i = 40; i >= 0; i--) {
      if ((l & (1LL << i)) > 0) {
        if (tmp + (1LL << i) > r) {
          ok = false;
        } else {
          a |= (1LL << i);
          b |= (l ^ (1LL << i));
        }
        break;
      }
    }
    if (ok) {
      cout << 0 << "\n" << 3 << "\n" << l << " " << a << " " << b << endl;
      return 0;
    }
  }
  if (k >= 4) {
    while (l + 3 <= r and mn > 0) {
      long long tmp = (l ^ (l + 1) ^ (l + 2) ^ (l + 3));
      if (tmp < mn) {
        mn = tmp;
        res.clear();
        res.push_back(l);
        res.push_back(l + 1);
        res.push_back(l + 2);
        res.push_back(l + 3);
      }
      l++;
    }
  }
  cout << mn << "\n" << res.size() << "\n";
  for (__typeof(res.begin()) it(res.begin()); it != res.end(); it++)
    cout << *it << " ";
  cout << endl;
  return 0;
}