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;
using ll = long long;
ll l, r, k;
bool f1(ll k) {
  if (k != 1) return false;
  printf("%I64d\n1\n%I64d\n", l, l);
  return true;
}
bool f2(ll k) {
  if (k != 2) return false;
  if (l + 1 == r && l < (l ^ r)) return f1(1);
  if ((l & 1) && l + 1 < r) ++l;
  printf("%I64d\n2\n%I64d %I64d\n", l ^ (l + 1), l, l + 1);
  return true;
}
bool f3(ll k) {
  if (k != 3) return false;
  for (ll p1 = 1; p1 <= r; p1 <<= 1) {
    ll p2 = p1 << 1;
    if (l < p2 && p1 + p2 <= r) {
      printf("0\n3\n%I64d %I64d %I64d\n", p2 - 1, p1 + p2 - 1, p1 + p2);
      return true;
    }
  }
  return f2(2);
}
bool f4(ll k) {
  if (k < 4) return false;
  for (ll i = l; i + 2 < r; ++i)
    if (!(i & 3) || !((i & 3) - 2)) {
      printf("0\n4\n%I64d %I64d %I64d %I64d\n", i, i + 1, i + 2, i + 3);
      return true;
    }
  return f3(3);
}
int main() {
  scanf("%I64d%I64d%I64d", &l, &r, &k);
  return f1(k) || f2(k) || f3(k) || f4(k), 0;
}