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, d;
pair<long long, long long> two() {
  if (d == 2) return make_pair(l ^ r, l);
  long long a = l;
  if (l & 1) a++;
  return make_pair(1, a);
}
pair<long long, long long> three() {
  long long t = 1;
  while (t <= l) t <<= 1;
  t |= (t >> 1);
  if ((t <= r) && (l ^ t) <= r) return make_pair(l, t);
  return make_pair(0, 0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> l >> r >> k;
  d = r - l + 1;
  pair<long long, long long> pp, qq;
  long long ans = l;
  if (k == 1) {
    cout << l << endl;
    cout << 1 << endl;
    cout << l << endl;
  } else if (k == 2) {
    pp = two();
    if (pp.first < l) {
      cout << pp.first << endl;
      cout << 2 << endl;
      cout << pp.second << ' ' << pp.second + 1 << endl;
    } else {
      cout << l << endl;
      cout << 1 << endl;
      cout << l << endl;
    }
  } else {
    pp = two();
    qq = three();
    if (qq.first > 0) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << qq.first << ' ' << qq.second << ' ' << (qq.first ^ qq.second)
           << endl;
    } else {
      if ((k == 3) or ((d == 4) && (l & 1))) {
        cout << 1 << endl;
        cout << 2 << endl;
        cout << pp.second << ' ' << pp.second + 1 << endl;
      } else {
        if (l & 1) l++;
        cout << 0 << endl;
        cout << 4 << endl;
        cout << l << ' ' << l + 1 << ' ' << l + 2 << ' ' << l + 3 << endl;
      }
    }
  }
}