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;
vector<long long> ans, seq;
int high1(long long num) {
  long long cnt = 0, lvl = 1;
  while (num >= lvl) lvl <<= 1, cnt++;
  return cnt;
}
int main() {
  long long l, r, k;
  ans.clear();
  scanf("%I64d%I64d%I64d", &l, &r, &k);
  if (k == 1)
    ans.push_back(l);
  else if (k == 2) {
    if (r - l + 1 == 2) {
      if ((l ^ r) < l)
        ans.push_back(l), ans.push_back(l + 1);
      else
        ans.push_back(l);
    } else {
      if (l & 1) l++;
      ans.push_back(l), ans.push_back(l + 1);
    }
  } else if (k == 3) {
    bool flg = false;
    for (long long i = 1; i <= r; i <<= 1)
      if (i >= l) {
        long long delt = (1LL << high1(r - i)) - 1;
        long long range = min(i - 1, delt);
        if (l <= range) {
          flg = true;
          if (i - 1 >= delt) {
            ans.push_back(i + (1LL << (high1(r - i) - 1)));
            ans.push_back(i + (1LL << (high1(r - i) - 1)) - 1);
            ans.push_back(range);
          } else {
            ans.push_back(i + i - 1);
            ans.push_back(i);
            ans.push_back(i - 1);
          }
          break;
        }
      }
    if (!flg) {
      if (l & 1) l++;
      ans.push_back(l), ans.push_back(l + 1);
    }
  } else {
    if (r - l + 1 >= 7) {
      for (; l & 3; l++)
        ;
      ans.push_back(l);
      ans.push_back(l + 1);
      ans.push_back(l + 2);
      ans.push_back(l + 3);
    } else {
      seq.clear();
      for (long long i = l; i <= r; i++) seq.push_back(i);
      int len = r - l;
      int lim = 1 << (r - l + 1);
      long long cans = 0x3f3f3f3f3f3f3f3fLL, cur, rec;
      for (int s = 1; s < lim; s++) {
        cur = 0;
        for (int i = 0; i <= len; i++)
          if (s & (1 << i)) cur ^= seq[i];
        if (cur < cans) cans = cur, rec = s;
      }
      for (int i = 0; i <= len; i++)
        if (rec & (1 << i)) ans.push_back(seq[i]);
    }
  }
  long long val = 0;
  for (int i = 0; i < (int)ans.size(); i++) val ^= ans[i];
  printf("%I64d\n%d\n", val, ans.size());
  std ::sort(ans.begin(), ans.end());
  for (int i = 0; i < (int)ans.size(); i++)
    printf("%I64d%c", ans[i], (i == (int)ans.size() - 1) ? '\n' : ' ');
}