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;
int main() {
  long long l, r, k;
  cin >> l >> r >> k;
  if (l == r || k == 1) {
    cout << l << "\n1\n" << l << "\n";
    return 0;
  }
  if (l + 1 == r) {
    if (l <= (l ^ r)) {
      cout << l << "\n1\n" << l << "\n";
    } else {
      cout << (l ^ r) << "\n2\n" << l << " " << r << "\n";
    }
    return 0;
  }
  if (k == 2) {
    if (l % 2) {
      cout << "1\n2\n" << l + 1 << " " << l + 2 << "\n";
    } else {
      cout << "1\n2\n" << l << " " << l + 1 << "\n";
    }
    return 0;
  }
  if (k == 3) {
    if (l == 1) {
      cout << "0\n3\n1 2 3\n";
      return 0;
    }
    vector<long long> p2(60, 1);
    for (int i = 1; i < 60; ++i) p2[i] = p2[i - 1] * 2;
    long long x1 = p2[upper_bound(p2.begin(), p2.end(), l) - p2.begin()],
              x2 = x1 / 2;
    if (x1 + x2 <= r) {
      cout << "0\n3\n" << l << " " << l + x1 - x2 << " " << x1 + x2 << "\n";
    } else {
      if (l % 2) {
        cout << "1\n2\n" << l + 1 << " " << l + 2 << "\n";
      } else {
        cout << "1\n2\n" << l << " " << l + 1 << "\n";
      }
    }
    return 0;
  }
  if (k >= 4) {
    r = min(r, l + 5);
    vector<long long> mask(1 << ((int)(r - l + 1)), 1e+18);
    vector<int> c(1 << ((int)(r - l + 1)), 0);
    for (long long i = l; i <= r; ++i) {
      mask[1 << ((int)(i - l))] = i;
      c[1 << ((int)(i - l))] = 1;
    }
    for (int i = 1; i < mask.size(); ++i) {
      for (int j = 0; j <= ((int)(r - l)); ++j) {
        if (!(i & (1 << j))) {
          if (c[i] < k) {
            mask[i | (1 << j)] =
                min(mask[i | (1 << j)], mask[i] ^ mask[1 << j]);
            c[i | (1 << j)] = c[i] + 1;
          }
        }
      }
    }
    long long res = 1e+18;
    int min_mask = -1;
    for (int i = 0; i < mask.size(); ++i) {
      if (mask[i] < res) {
        res = mask[i];
        min_mask = i;
      }
    }
    cout << res << "\n";
    vector<long long> ans(0);
    for (long long i = l; i <= r; ++i)
      if (min_mask & (1 << ((int)(i - l)))) ans.push_back(i);
    cout << ans.size() << "\n" << ans[0];
    for (int i = 1; i < ans.size(); ++i) cout << " " << ans[i];
    cout << "\n";
  }
  return 0;
}