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;
int main() {
  cin >> l >> r >> k;
  long long i;
  if (k >= 4) {
    for (i = l; i + 3 <= r; i++) {
      if ((i & 1) == 0) {
        printf("0\n4\n%I64d %I64d %I64d %I64d\n", i, i + 1, i + 2, i + 3);
        return 0;
      }
    }
  }
  if (k >= 3) {
    long long t;
    t = l;
    i = 0;
    while (t > 0) {
      i++;
      t >>= 1;
    }
    t = 3;
    while (--i) {
      t <<= 1;
    }
    if (t <= r) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << l << ' ' << (l ^ t) << ' ' << t << endl;
      return 0;
    }
  }
  if (k >= 2) {
    for (i = l; i + 1 <= r; i++) {
      if ((i & 1) == 0) {
        cout << 1 << endl;
        cout << 2 << endl;
        cout << i << ' ' << i + 1 << endl;
        return 0;
      }
    }
    if ((l ^ r) < l) {
      printf("%I64d\n2\n%I64d %I64d\n", l ^ r, l, r);
      return 0;
    }
  }
  printf("%I64d\n1\n%I64d", l, l);
  return 0;
}