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;
void go(long long x) {
  int i, j;
  long long y, z;
  for (i = 62; i >= 0; i--) {
    if (x & (1LL << i)) break;
  }
  if ((1LL << i) == x) {
    x++;
  }
  long long xx = x;
  xx &= ((1LL << i) - 1);
  y = (1LL << (i + 1)) | xx;
  z = (1LL << (i + 1)) | (1LL << i);
  if (y <= r && z <= r && k >= 3) {
    cout << 0 << endl << 3 << endl << x << ' ' << y << ' ' << z << endl;
    exit(0);
  }
}
int main() {
  cin >> l >> r >> k;
  pair<long long, long long> p;
  if (l == 1) {
    if (k >= 3) {
      cout << 0 << endl << 3 << endl;
      cout << 1 << ' ' << 2 << ' ' << 3 << endl;
      return 0;
    } else {
      cout << 1 << endl << 1 << endl << 1 << endl;
      return 0;
    }
  }
  go(l);
  if (k >= 5) {
    cout << 0 << endl << 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 if (k == 4) {
    if (l % 2 == 0) {
      cout << 0 << endl << 4 << endl;
      cout << l << ' ' << l + 1 << ' ' << l + 2 << ' ' << l + 3 << endl;
    } else {
      if (r - l + 1 >= 5) {
        cout << 0 << endl << 4 << endl;
        cout << l + 1 << ' ' << l + 2 << ' ' << l + 3 << ' ' << l + 4 << endl;
      } else {
        cout << 1 << endl << 2 << endl;
        if (l % 2 == 0)
          cout << l << ' ' << l + 1 << endl;
        else
          cout << l + 1 << ' ' << l + 2 << endl;
      }
    }
  } else if (k == 3) {
    cout << 1 << endl << 2 << endl;
    if (l % 2 == 0)
      cout << l << ' ' << l + 1 << endl;
    else
      cout << l + 1 << ' ' << l + 2 << endl;
  } else if (k == 2) {
    if (r - l + 1 > 2) {
      cout << 1 << endl << 2 << endl;
      if (l % 2 == 0)
        cout << l << ' ' << l + 1 << endl;
      else
        cout << l + 1 << ' ' << l + 2 << endl;
    } else {
      if ((l ^ (l + 1)) < l) {
        cout << (l ^ (l + 1)) << endl << 2 << endl;
        cout << l << ' ' << l + 1 << endl;
      } else
        cout << l << endl << 1 << endl << l << endl;
    }
  } else {
    cout << l << endl << 1 << endl << l << endl;
  }
}