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;
bool PIIfs(const pair<int, int>& a, const pair<int, int>& b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
bool PIIsf(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second < b.second || (a.second == b.second && a.first < b.first);
}
bool PIIFS(const pair<int, int>& a, const pair<int, int>& b) {
  return a.first > b.first || (a.first == b.first && a.second > b.second);
}
bool PIISF(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second > b.second || (a.second == b.second && a.first > b.first);
}
bool PIIfS(const pair<int, int>& a, const pair<int, int>& b) {
  return a.first < b.first || (a.first == b.first && a.second > b.second);
}
bool PIIsF(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second < b.second || (a.second == b.second && a.first > b.first);
}
bool PIIFs(const pair<int, int>& a, const pair<int, int>& b) {
  return a.first > b.first || (a.first == b.first && a.second < b.second);
}
bool PIISf(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second > b.second || (a.second == b.second && a.first < b.first);
}
template <typename T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <typename T>
inline T max(T a, T b, T c, T d) {
  return max(max(a, b, c), d);
}
template <typename T>
inline T max(T a, T b, T c, T d, T e) {
  return max(max(a, b, c, d), e);
}
template <typename T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <typename T>
inline T min(T a, T b, T c, T d) {
  return min(min(a, b, c), d);
}
template <typename T>
inline T min(T a, T b, T c, T d, T e) {
  return min(min(a, b, c, d), e);
}
template <typename T>
inline istream& RD(T& x) {
  return cin >> x;
}
template <typename T, typename U>
inline istream& RD(T& a, U& b) {
  return cin >> a >> b;
}
template <typename T, typename U, typename V>
inline istream& RD(T& a, U& b, V& c) {
  return cin >> a >> b >> c;
}
template <typename T, typename U, typename V, typename W>
inline istream& RD(T& a, U& b, V& c, W& d) {
  return cin >> a >> b >> c >> d;
}
template <typename T, typename U, typename V, typename W, typename X>
inline istream& RD(T& a, U& b, V& c, W& d, X& e) {
  return cin >> a >> b >> c >> d >> e;
}
template <typename T>
void checkmin(T& a, const T& b) {
  if (b < a) a = b;
}
template <typename T>
void checkmax(T& a, const T& b) {
  if (b > a) a = b;
}
long long msb(long long x) {
  long long ans = 1;
  while ((x >>= 1)) ans <<= 1;
  return ans;
}
int main() {
  std::ios::sync_with_stdio(false);
  long long l, r, k;
  cin >> l >> r >> k;
  if (r - l + 1 >= 7 && k >= 4) {
    cout << "0\n";
    while (l % 4 != 0) l++;
    cout << "4\n";
    cout << l << ' ' << l + 1 << ' ' << l + 2 << ' ' << l + 3 << '\n';
  } else if (r - l + 1 < 7) {
    int bits = r - l + 1;
    long long minans = l;
    for (int i = 1; i < (1 << bits); i++) {
      long long tempxor = 0;
      bitset<6> bs(i);
      if (bs.count() > k) continue;
      for (int b = 0; b < 6; b++) {
        if (bs.test(b)) {
          tempxor ^= (l + b);
        }
      }
      checkmin(minans, tempxor);
    }
    for (int i = 1; i < (1 << bits); i++) {
      long long tempxor = 0;
      bitset<6> bs(i);
      if (bs.count() > k) continue;
      for (int b = 0; b < 6; b++) {
        if (bs.test(b)) {
          tempxor ^= (l + b);
        }
      }
      if (tempxor == minans) {
        cout << tempxor << '\n';
        cout << bs.count() << '\n';
        for (int b = 0; b < 5; b++) {
          if (bs.test(b)) {
            cout << l + b << ' ';
          }
        }
        cout << '\n';
        break;
      }
    }
  } else if (k == 3) {
    if (msb(l) == msb(r)) {
      if ((l ^ (l + 1)) == 1) {
        r = l + 1;
      } else {
        l++;
        r = l + 1;
      }
      cout << (l ^ r) << '\n';
      cout << "2\n";
      cout << l << ' ' << r << '\n';
    } else {
      long long ones = (msb(l) << 1) - 1;
      long long sb = ones + 1 + ((ones + 1) >> 1);
      if (r >= sb) {
        cout << "0\n";
        cout << "3\n";
        cout << sb << ' ' << sb - 1 << ' ' << ones << '\n';
      } else {
        if ((l ^ (l + 1)) == 1) {
          r = l + 1;
        } else {
          l++;
          r = l + 1;
        }
        cout << (l ^ r) << '\n';
        cout << "2\n";
        cout << l << ' ' << r << '\n';
      }
    }
  } else if (k == 2) {
    if (r - l == 1) {
      if ((l ^ r) < l) {
        cout << (l ^ r) << '\n';
        cout << "2\n";
        cout << l << ' ' << r << '\n';
      } else {
        cout << l << '\n' << 1 << '\n' << l << '\n';
      }
    } else {
      if ((l ^ (l + 1)) == 1) {
        r = l + 1;
      } else {
        l++;
        r = l + 1;
      }
      cout << (l ^ r) << '\n';
      cout << "2\n";
      cout << l << ' ' << r << '\n';
    }
  } else {
    cout << l << '\n' << "1\n" << l << '\n';
  }
  return 0;
}