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 take2(long long l, long long r) {
  if (r - l == 1) {
    if ((l ^ (l + 1)) > l) {
      cout << l << endl;
      cout << "1" << endl;
      cout << l << endl;
    } else {
      cout << (l ^ (l + 1)) << endl;
      cout << "2" << endl;
      cout << l << " " << l + 1 << endl;
    }
  } else {
    if (l % 2) l++;
    cout << (l ^ (l + 1)) << endl;
    cout << "2" << endl;
    cout << l << " " << l + 1 << endl;
  }
}
void take3(long long L, long long R) {
  for (int bit = 1; bit <= 60; ++bit) {
    long long int up = (1LL << bit) + (1LL << (bit - 1));
    long long int middle = up - 1;
    long long int low = (1LL << bit) - 1;
    if (L <= low && up <= R) {
      cout << "0" << endl;
      cout << "3" << endl;
      cout << low << " " << middle << " " << up << endl;
      return;
    }
  }
  take2(L, R);
}
void take4(long long L, long long R) {
  if (R - L > 3 || L % 2 == 0) {
    if (L % 2) ++L;
    cout << "0" << endl;
    cout << "4" << endl;
    for (long long i = L; i <= L + 3; ++i) cout << i << " ";
  } else
    take3(L, R);
}
int main() {
  long long L, R;
  int maxSize;
  cin >> L >> R >> maxSize;
  if (maxSize == 1) {
    cout << L << endl;
    cout << "1" << endl;
    cout << L << endl;
  }
  if (maxSize == 2) take2(L, R);
  if (maxSize == 3) take3(L, R);
  if (maxSize >= 4) take4(L, R);
  return 0;
}