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>
const int MAXN = 2e5 + 15;
using namespace std;
pair<long long, long long> solvefor3(long long l, long long r) {
  long long x = 1;
  int m = 0;
  while (x < l) {
    m++;
    x *= 2;
    x++;
  }
  long long y = (1LL << m) + x;
  if ((y ^ x) <= r)
    return {x, y};
  else
    return {-1, -1};
}
void solve() {
  long long l, r, k;
  cin >> l >> r >> k;
  if (k == 1) {
    cout << l << "\n1\n" << l;
    return;
  }
  if (k == 2) {
    if (r == l + 1) {
      if (l < (l ^ r)) {
        cout << l << "\n1\n" << l;
        return;
      }
      cout << (l ^ r) << "\n2\n" << l << " " << r;
    } else {
      if (l & 1) l++;
      cout << 1 << "\n";
      cout << 2 << "\n";
      cout << l << " " << l + 1;
    }
    return;
  }
  auto pp = solvefor3(l, r);
  if (k == 3) {
    if (pp.first != -1) {
      cout << "0\n3\n"
           << pp.first << " " << pp.second << " " << (pp.first ^ pp.second);
      return;
    } else {
      if (l & 1) l++;
      cout << 1 << "\n";
      cout << 2 << "\n";
      cout << l << " " << l + 1;
    }
  }
  if (k >= 4) {
    if (r - l >= 4 || (l & 1) == 0) {
      if (l & 1) l++;
      cout << 0 << "\n4\n";
      cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3;
      return;
    } else {
      if (pp.first != -1) {
        cout << "0\n3\n"
             << pp.first << " " << pp.second << " " << (pp.first ^ pp.second);
        return;
      } else {
        if (l & 1) l++;
        cout << 1 << "\n";
        cout << 2 << "\n";
        cout << l << " " << l + 1;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cout << fixed << setprecision(12);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}