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;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
long long l, r, a, b, c;
int k;
bool find3() {
  a = 1;
  while (a < l) a <<= 1, a |= 1;
  b = (a << 1) ^ a ^ 1 ^ (a >> 1);
  c = a ^ b;
  return c <= r;
}
int main() {
  read(l);
  read(r);
  read(k);
  if (k == 1)
    printf("%lld\n1\n%lld\n", l, l);
  else if (k == 2) {
    if (!(l & 1))
      printf("1\n2\n%lld %lld\n", l, l + 1);
    else if (r - l != 1)
      printf("1\n2\n%lld %lld\n", l + 1, l + 2);
    else if ((l ^ (l + 1)) < l)
      printf("%lld\n2\n%lld\n%lld\n", l ^ (l + 1), l, l + 1);
    else
      printf("%lld\n1\n%lld\n", l, l);
  } else if (k == 3) {
    if (find3())
      printf("0\n3\n%lld %lld %lld\n", a, b, c);
    else if (!(l & 1))
      printf("1\n2\n%lld %lld\n", l, l + 1);
    else
      printf("1\n2\n%lld %lld\n", l + 1, l + 2);
  } else if (k >= 4) {
    if (!(l & 1))
      printf("0\n4\n%lld %lld %lld %lld\n", l, l + 1, l + 2, l + 3);
    else if (r - l != 3)
      printf("0\n4\n%lld %lld %lld %lld\n", l + 1, l + 2, l + 3, l + 4);
    else if (find3())
      printf("0\n3\n%lld %lld %lld\n", a, b, c);
    else if (!(l & 1))
      printf("1\n2\n%lld %lld\n", l, l + 1);
    else
      printf("1\n2\n%lld %lld\n", l + 1, l + 2);
  }
  return 0;
}