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;
void search(long long L, long long R, long long K) {
  long long range = R - L + 1;
  long long ret_value = L;
  long long ret_cnt = 0;
  vector<long long> ret_nums;
  for (int code = 1; code < (1 << range); ++code) {
    long long value = 0;
    long long cnt = 0;
    vector<long long> nums;
    for (int i = 0; i < range; ++i)
      if ((code & (1 << i)) != 0) {
        value ^= L + i;
        cnt++;
        nums.push_back(L + i);
      }
    if (value <= ret_value && cnt >= 1 && cnt <= K) {
      ret_value = value;
      ret_cnt = cnt;
      ret_nums.assign(nums.begin(), nums.end());
    }
  }
  cout << ret_value << endl;
  cout << ret_cnt << endl;
  for (int i = 0; i < ret_cnt; ++i)
    cout << ret_nums[i] << (i == ret_cnt - 1 ? '\n' : ' ');
}
void answer1(long long L) {
  cout << L << endl;
  cout << 1 << endl;
  cout << L << endl;
}
void answer2(long long L) {
  cout << 1 << endl;
  cout << 2 << endl;
  if ((L & 1) == 0)
    cout << L << ' ' << L + 1;
  else
    cout << L + 1 << ' ' << L + 2 << endl;
}
void answer3(long long L, long long R) {
  for (long long i = 1;; i++) {
    long long a = (1LL << i) + (1LL << (i - 1));
    long long b = (1LL << i) - 1;
    long long c = (1LL << i) + (1LL << (i - 1)) - 1;
    if (a >= L && a <= R && b >= L && b <= R && c >= L && c <= R) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << a << ' ' << b << ' ' << c << ' ' << endl;
      return;
    }
    if (a > R || b > R || c > R) break;
  }
  answer2(L);
}
void answer4(long long L) {
  cout << 0 << endl;
  cout << 4 << endl;
  if ((L & 1) == 0)
    cout << L << ' ' << L + 1 << ' ' << L + 2 << ' ' << L + 3 << endl;
  else
    cout << L + 1 << ' ' << L + 2 << ' ' << L + 3 << ' ' << L + 4 << endl;
}
int main() {
  long long L, R, K;
  cin >> L >> R >> K;
  long long range = R - L + 1;
  if (range < 5)
    search(L, R, K);
  else if (K == 1)
    answer1(L);
  else if (K == 2)
    answer2(L);
  else if (K == 3)
    answer3(L, R);
  else
    answer4(L);
  return 0;
}