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;
unsigned long long l, r, k, sol[4], best[4], bestXor = 1e12 + 1, bestN = 0;
void brute(unsigned long long num, unsigned long long start,
           unsigned long long end, unsigned long long Xor) {
  if (num > k) return;
  if (num > 0 && bestXor > Xor) {
    for (__typeof(4) i = 0; i < 4; ++i) best[i] = sol[i];
    bestXor = Xor;
    bestN = num;
  }
  for (__typeof(end) i = start; i <= end; ++i) {
    sol[num] = i;
    brute(num + 1, i + 1, end, Xor ^ i);
  }
}
unsigned long long hob(unsigned long long num) {
  if (!num) return 0;
  unsigned long long ret = 1;
  while (num >>= 1) ret <<= 1;
  return ret;
}
void doFirst(int n) {
  brute(0, l, min(r, l + n - 1), 0);
  cout << bestXor << endl;
  cout << bestN << endl;
  for (__typeof(bestN) i = 0; i < bestN; ++i) cout << best[i] << " ";
}
int main() {
  cin >> l >> r >> k;
  if (k == 1 || l == r) {
    cout << l << endl << 1 << endl << l;
  } else if (k == 2) {
    doFirst(3);
  } else if (k == 3) {
    unsigned long long h = hob(r);
    if ((r ^ h) >= l) {
      cout << 0 << endl << 3 << endl;
      cout << (r ^ h) << " " << r << " " << h << endl;
    } else if (((h - 1) >= (l) && (h - 1) <= (r)) &&
               ((h + h / 2) >= (l) && (h + h / 2) <= (r)) &&
               ((h + h / 2 - 1) >= (l) && (h + h / 2 - 1) <= (r))) {
      cout << 0 << endl << 3 << endl;
      cout << (h - 1) << " " << (h + h / 2) << " " << (h + h / 2 - 1) << endl;
    } else {
      doFirst(3);
    }
  } else {
    doFirst(5);
  }
  return 0;
}