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;
int32_t main() {
  scanf("%lld%lld%lld", &l, &r, &k);
  if (r - l + 1 <= 5) {
    pair<long long, long long> Mn = {LLONG_MAX, 0};
    for (long long i = 1; i < (1 << r - l + 1); i++)
      if (__builtin_popcountll(i) <= k) {
        long long res = 0;
        for (long long j = 0; j < r - l + 1; j++)
          if (i & (1 << j)) res ^= (l + j);
        Mn = min(Mn, {res, i});
      }
    printf("%lld\n%d\n", Mn.first, __builtin_popcountll(Mn.second));
    for (long long j = 0; j < r - l + 1; j++)
      if (Mn.second & (1 << j)) printf("%lld ", l + j);
    return (0);
  }
  if (k == 1) return !printf("%lld\n1\n%lld", l, l);
  if (k != 3 && (l & 1)) l++;
  if (k == 2) return !printf("1\n2\n%lld %lld", l, l + 1);
  if (k >= 4)
    return !printf("0\n4\n%lld %lld %lld %lld", l, l + 1, l + 2, l + 3);
  for (long long j = 50; j; j--)
    if ((1ll << j) + (1ll << j - 1) <= r && l <= (1ll << j) - 1)
      return !printf("0\n3\n%lld %lld %lld", (1ll << j) + (1ll << j - 1),
                     (1ll << j) + (1ll << j - 1) - 1, (1ll << j) - 1);
  if (l & 1) l++;
  return !printf("1\n2\n%lld %lld", l, l + 1);
}