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;
const int BigInt = 2000 * 1000 * 1000;
long long l, r, k;
long long ans = 1ll << 60;
vector<long long> ansVector;
void bruteforce(long long number, long long sum,
                const vector<long long>& elems) {
  if (number > r) {
    if (1 <= elems.size() && elems.size() <= k) {
      if (sum < ans) {
        ans = sum;
        ansVector = elems;
      }
    }
    return;
  }
  bruteforce(number + 1, sum, elems);
  vector<long long> elemsAndCur = elems;
  elemsAndCur.push_back(number);
  bruteforce(number + 1, sum ^ number, elemsAndCur);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> l >> r >> k;
  if (k == 3) {
    for (int b1 = 0; b1 < 60; ++b1) {
      long long low = (1ll << (b1 + 1)) - 1;
      long long high = (3ll << b1);
      if ((l <= low && low <= r) && (l <= high && high <= r)) {
        long long middle = low ^ high;
        cout << 0 << endl;
        cout << 3 << endl;
        cout << low << ' ' << high << ' ' << middle << endl;
        return 0;
      }
    }
  }
  r = min(r, l + 7);
  vector<long long> elems;
  bruteforce(l, 0, elems);
  cout << ans << endl;
  cout << ansVector.size() << endl;
  for (long long v : ansVector) cout << v << ' ';
  cout << endl;
  return 0;
}