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;
inline string GetString() {
  char GS[1000005];
  scanf("%s", GS);
  string ret = GS;
  return ret;
}
inline char getc() {
  char c = ' ';
  while (c == ' ' || c == '\t' || c == '\r' || c == '\n') c = getchar();
  return c;
}
long long mini;
vector<long long> nummin;
long long l, r, k;
int main() {
  mini = 1000000000000000LL;
  scanf("%I64d %I64d %I64d", &l, &r, &k);
  if (k >= 1) {
    if (l < mini) {
      mini = l;
      nummin = {l};
    }
  }
  if (k >= 2) {
    if ((l ^ (l + 1)) < mini) {
      mini = l ^ (l + 1);
      nummin = {l, l + 1};
    }
    if (l + 2 <= r && ((l + 1) ^ (l + 2)) < mini) {
      mini = (l + 1) ^ (l + 2);
      nummin = {l + 1, l + 2};
    }
  }
  if (k >= 3) {
    for (long long i = 1; i <= 45; i++) {
      long long a = (1LL << i) + (1LL << (i - 1));
      long long b = a - 1;
      long long c = (1LL << i) - 1;
      if (l <= c && a <= r) {
        if ((a ^ b ^ c) < mini) {
          mini = a ^ b ^ c;
          nummin = {a, b, c};
        }
      }
    }
  }
  if (k >= 4) {
    if ((l ^ (l + 1) ^ (l + 2) ^ (l + 3)) < mini) {
      mini = l ^ (l + 1) ^ (l + 2) ^ (l + 3);
      nummin = {l, l + 1, l + 2, l + 3};
    }
    if (l + 4 <= r && ((l + 1) ^ (l + 2) ^ (l + 3) ^ (l + 4)) < mini) {
      mini = (l + 1) ^ (l + 2) ^ (l + 3) ^ (l + 4);
      nummin = {l + 1, l + 2, l + 3, l + 4};
    }
  }
  printf("%I64d\n", mini);
  printf("%d\n", nummin.size());
  for (auto a : nummin) printf("%I64d ", a);
}