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() {
  ios_base::sync_with_stdio(0);
  long long int l, r, k;
  cin >> l >> r >> k;
  ;
  if (k == 1) {
    cout << l << endl << 1 << endl << l << endl;
  }
  if (k == 2) {
    if (r - l == 1) {
      if (l < (l ^ r))
        cout << l << endl << 1 << endl << l << endl;
      else
        cout << (l ^ r) << endl << 2 << endl << l << " " << r << endl;
    } else {
      cout << 1 << endl;
      cout << 2 << endl;
      if (l % 2 == 0)
        cout << l << " " << l + 1 << endl;
      else
        cout << l + 1 << " " << l + 2 << endl;
    }
  }
  if (k == 3) {
    long long int x = 1LL;
    while (x <= l) x *= 2;
    long long int y = x / 2;
    if (x + y <= r) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << l << " " << x + (l - y) << " " << x + y << endl;
    } else {
      cout << 1 << endl;
      cout << 2 << endl;
      if (l % 2 == 0)
        cout << l << " " << l + 1 << endl;
      else
        cout << l + 1 << " " << l + 2 << endl;
    }
  }
  if (k == 4) {
    if (r - l > 3) {
      cout << "0" << endl;
      cout << "4" << endl;
      if (l % 2 == 0)
        cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
      else
        cout << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4 << endl;
    } else {
      long long int best = r;
      vector<long long int> dla;
      for (int i = (1); i < (16); ++i) {
        int mask = i;
        vector<long long int> v;
        int step = 0;
        while (mask != 0) {
          if (mask % 2) v.push_back(l + step);
          step++;
          mask /= 2;
        }
        long long int wyn = v[0];
        for (int i = (1); i < (v.size()); ++i) wyn = wyn ^ v[i];
        if (wyn < best) {
          best = wyn;
          dla = v;
        }
      }
      cout << best << endl;
      cout << dla.size() << endl;
      for (int i = (0); i < (dla.size()); ++i) cout << dla[i] << " ";
      cout << endl;
    }
  }
  if (k > 4) {
    cout << "0" << endl;
    cout << "4" << endl;
    if (l % 2 == 0)
      cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
    else
      cout << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4 << endl;
  }
  return 0;
}