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;
long long L, R, K;
long long vio[5];
inline void scanff() { cin >> L >> R >> K; }
inline void violence() {
  if (K == 1) {
    cout << L << endl << 1 << endl << L << endl;
  }
  if (K == 2) {
    if (R - L + 1 == 2) {
      if ((L ^ R) < L)
        cout << (L ^ L + 1) << endl << 2 << endl << L << " " << R << endl;
      else
        cout << L << endl << 1 << endl << L << endl;
    } else {
      for (long long i = L; i <= R; ++i)
        if (i % 2 == 0) {
          cout << 1 << endl << 2 << endl << i << " " << i + 1 << endl;
          break;
        }
    }
  }
  if (K == 3) {
    long long a = 3, b = 2, c = 1;
    while ((a << 1) <= R && ((b << 1) | 1) <= R && ((c << 1) | 1) << R &&
           a <= L && b <= L && c <= L) {
      a = (a << 1);
      b = ((b << 1) | 1);
      c = ((c << 1) | 1);
    }
    if (a >= L && a <= R && b >= L && b <= R && c >= L && c <= R)
      cout << 0 << endl << 3 << endl << c << " " << b << " " << a << endl;
    else {
      for (long long i = L; i <= R; ++i)
        if (i % 2 == 0) {
          cout << 1 << endl << 2 << endl << i << " " << i + 1 << endl;
          break;
        }
    }
  }
  if (K == 4) {
    if (L % 2 == 0) {
      cout << 0 << endl
           << 4 << endl
           << L << " " << L + 1 << " " << L + 2 << " " << L + 3 << endl;
    } else {
      long long a = 3, b = 2, c = 1;
      while ((a << 1) <= R && ((b << 1) | 1) <= R && ((c << 1) | 1) << R &&
             a <= L && b <= L && c <= L) {
        a = (a << 1);
        b = ((b << 1) | 1);
        c = ((c << 1) | 1);
      }
      if (a >= L && a <= R && b >= L && b <= R && c >= L && c <= R)
        cout << 0 << endl << 3 << endl << c << " " << b << " " << a << endl;
      else {
        for (long long i = L; i <= R; ++i)
          if (i % 2 == 0) {
            cout << 1 << endl << 2 << endl << i << " " << i + 1 << endl;
            break;
          }
      }
    }
  }
}
inline void solve() {
  if (K == 1)
    cout << L << endl << 1 << endl << L << endl;
  else if (K == 2) {
    for (long long i = L; i <= R; ++i)
      if (i % 2 == 0) {
        cout << 1 << endl << 2 << endl << i << " " << i + 1 << endl;
        break;
      }
  } else if (K == 3) {
    long long a = 3, b = 2, c = 1;
    while ((a << 1) <= R && ((b << 1) | 1) <= R && ((c << 1) | 1) << R &&
           c <= L) {
      a = (a << 1);
      b = ((b << 1) | 1);
      c = ((c << 1) | 1);
    }
    if (a >= L && a <= R && b >= L && b <= R && c >= L && c <= R)
      cout << 0 << endl << 3 << endl << c << " " << b << " " << a << endl;
    else {
      for (long long i = L; i <= R; ++i)
        if (i % 2 == 0) {
          cout << 1 << endl << 2 << endl << i << " " << i + 1 << endl;
          break;
        }
    }
  } else if (K == 4) {
    for (long long i = L; i <= R; ++i)
      if (i % 2 == 0) {
        cout << 0 << endl
             << 4 << endl
             << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << endl;
        break;
      }
  } else {
    for (long long i = L; i <= R; ++i)
      if (i % 2 == 0) {
        cout << 0 << endl
             << 4 << endl
             << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << endl;
        break;
      }
  }
}
inline void work() {
  if ((R - L + 1) <= 4) {
    violence();
    return;
  }
  solve();
}
int main() {
  scanff();
  work();
  return 0;
}