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;
inline bool CanBe0() {
  int Ca, Cb, ar[55], br[55], i, j;
  long long t, l, r;
  if (K < 3) return 0;
  for (Ca = 0, l = L; l; l >>= 1) ar[++Ca] = l & 1;
  for (Cb = 0, r = R; r; r >>= 1) br[++Cb] = r & 1;
  if ((Ca != Cb) && (K > 2)) {
    bool flag = false;
    for (i = Ca; i <= Cb - 1; ++i)
      if (br[i]) flag = true;
    if (flag) {
      long long t0 = (1LL << (Cb - 1)) + (1LL << (Ca - 1));
      long long t1 = (1LL << (Cb - 1)) + (L ^ (1LL << (Ca - 1)));
      cout << "0" << endl << "3" << endl << t0 << " " << t1 << " " << L << endl;
      return 1;
    }
  }
  if (K < 4) return 0;
  for (i = 0; i <= 3; ++i) {
    if (L + i + 3 <= R) {
      t = 0;
      for (j = 0; j <= 3; ++j) t ^= (L + i + j);
      if (!t) {
        cout << "0" << endl
             << "4" << endl
             << L + i << " " << L + i + 1 << " " << L + i + 2 << " "
             << L + i + 3 << endl;
        return 1;
      }
    }
    if (R - i - 3 >= L) {
      t = 0;
      for (j = 0; j <= 3; ++j) t ^= (R - i - j);
      if (!t) {
        cout << "0" << endl
             << "4" << endl
             << R - i << " " << R - i - 1 << " " << R - i - 2 << " "
             << R - i - 3 << endl;
        return 1;
      }
    }
  }
  return 0;
}
inline bool CanBe1() {
  int i, j;
  long long t;
  if (K < 2) return 0;
  for (i = 0; i <= 3; ++i) {
    if (L + i + 1 <= R) {
      t = 0;
      for (j = 0; j <= 1; ++j) t ^= (L + i + j);
      if (t == 1) {
        cout << "1" << endl << "2" << endl << L + i << " " << L + i + 1 << endl;
        return 1;
      }
    }
    if (R - i - 1 >= L) {
      t = 0;
      for (j = 0; j <= 1; ++j) t ^= (R - i - j);
      if (t == 1) {
        cout << "1" << endl << "2" << endl << R - i << " " << R - i - 1 << endl;
        return 1;
      }
    }
  }
  return 0;
}
inline void Calc() {
  int i;
  long long r, l;
  if (K == 1) {
    cout << L << endl << "1" << endl << L << endl;
    return;
  }
  if (L == R) {
    cout << L << endl << "1" << endl << L << endl;
    return;
  }
  if (CanBe0()) return;
  if (CanBe1()) return;
  if ((R ^ L) < L) {
    cout << (R ^ L) << endl << "2" << endl << R << " " << L << endl;
    return;
  } else {
    cout << L << endl << "1" << endl << L << endl;
    return;
  }
}
int main() {
  cin >> L >> R >> K;
  Calc();
  return 0;
}