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() {
  scanf("%lld%lld%lld", &l, &r, &k);
  if (k == 1ll)
    printf("%lld\n%lld\n%lld\n", l, 1ll, l);
  else if (k == 2ll) {
    if (l == 1ll)
      printf("%lld\n%lld\n%lld\n", 1ll, 1ll, l);
    else if (r - l == 1ll && (l & 1ll)) {
      long long ans = l ^ (l + 1);
      if (l < ans)
        printf("%lld\n%lld\n%lld\n", l, 1ll, l);
      else
        printf("%lld\n%lld\n%lld %lld\n", ans, 2ll, l, l + 1);
    } else if (l & 1ll)
      printf("%lld\n%lld\n%lld %lld\n", 1ll, 2ll, l + 1, l + 2);
    else
      printf("%lld\n%lld\n%lld %lld\n", 1ll, 2ll, l, l + 1);
  } else if (k == 3ll) {
    bool flag = false;
    for (register int i = 62; i >= 0; --i) {
      long long d1 = (1ll << i), d2 = (1ll << (i - 1));
      long long a = d1 | d2, b = d1 | (d2 - 1), c = d1 - 1;
      if ((l <= a && a <= r) && (l <= b && b <= r) && (l <= c && c <= r)) {
        flag = true;
        printf("%lld\n%lld\n%lld %lld %lld\n", 0ll, 3ll, a, b, c);
        break;
      }
    }
    if (!flag) {
      if (l & (1ll))
        printf("%lld\n%lld\n%lld %lld\n", 1ll, 2ll, l + 1, l + 2);
      else
        printf("%lld\n%lld\n%lld %lld\n", 1ll, 2ll, l, l + 1);
    }
  } else {
    if (r - l + 1 > 4) {
      if (l & 1ll)
        printf("%lld\n%lld\n%lld %lld %lld %lld\n", 0ll, 4ll, l + 1, l + 2,
               l + 3, l + 4);
      else
        printf("%lld\n%lld\n%lld %lld %lld %lld\n", 0ll, 4ll, l, l + 1, l + 2,
               l + 3);
    } else if (r - l + 1 == 4ll) {
      bool flag = false;
      for (register int i = 62; i >= 0; --i) {
        long long d1 = (1ll << i), d2 = (1ll << (i - 1));
        long long a = d1 | d2, b = d1 | (d2 - 1), c = d1 - 1;
        if ((l <= a && a <= r) && (l <= b && b <= r) && (l <= c && c <= r)) {
          flag = true;
          printf("%lld\n%lld\n%lld %lld %lld\n", 0ll, 3ll, a, b, c);
          break;
        }
      }
      if (!flag) {
        if (l & 1ll)
          printf("%lld\n%lld\n%lld %lld\n", 1ll, 2ll, l + 1, l + 2);
        else
          printf("%lld\n%lld\n%lld %lld %lld %lld\n", 0ll, 4ll, l, l + 1, l + 2,
                 l + 3);
      }
    }
  }
  return 0;
}