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;
const long long INF = 1000000000000000ll;
long long a, b, ans;
int k;
vector<long long> best, current;
void rec(long long x, long long res, bool empty) {
  if (x == b + 1) {
    if (empty) return;
    if (res < ans) {
      ans = res;
      best = current;
    }
    return;
  }
  if ((int)current.size() < k) {
    current.push_back(x);
    rec(x + 1, (res ^ x), 0);
    current.pop_back();
  }
  rec(x + 1, res, empty);
}
int main() {
  ans = INF;
  cin >> a >> b >> k;
  if (b - a <= 10) {
    rec(a, 0, 1);
    cout << ans << endl;
    cout << best.size() << endl;
    for (int i = 0; i < best.size(); i++) cout << best[i] << " ";
    return 0;
  }
  if (k == 1) {
    cout << a << endl;
    cout << 1 << endl;
    cout << a << endl;
    return 0;
  }
  if (k == 2) {
    cout << 1 << endl;
    if (a % 2 == 1) a++;
    cout << 2 << endl;
    cout << a << " " << a + 1;
    return 0;
  }
  if (k >= 4) {
    cout << 0 << endl;
    cout << 4 << endl;
    if (a % 2 == 1) a++;
    cout << a << " " << a + 1 << " " << a + 2 << " " << a + 3;
    return 0;
  }
  long long l = 1, r = 3;
  while (r <= b) {
    if (l >= a) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << l << " " << r - 1 << " " << r;
      return 0;
    }
    l *= 2;
    l++;
    r *= 2;
  }
  cout << 1 << endl;
  cout << 2 << endl;
  if (a % 2 == 1) a++;
  cout << a << " " << a + 1;
  return 0;
}