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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a = 0, b, c, d, e = 0, f = 0, g, h = 0, x, y, z;
  cin >> a >> b >> c;
  if (c == 1) {
    cout << a << endl;
    cout << 1 << endl;
    cout << a << endl;
    return 0;
  }
  if (c == 2) {
    d = 100000000000000ll;
    e = -1;
    for (long long i = a; i < b; i++) {
      f = i ^ (i + 1);
      if (f < d) {
        d = f;
        e = i;
      }
      if (f == 1) break;
    }
    if (d > a) {
      cout << a << endl << 1 << endl << a << endl;
      return 0;
    }
    cout << d << endl;
    cout << 2 << endl;
    cout << e << " " << e + 1 << endl;
    return 0;
  }
  if (c == 3) {
    for (long long i = a; i <= min(b, a + 100000ll); i++) {
      for (int j = 0; j < 50; j++) {
        if (i < 1ll << j) {
          d = 1ll << j;
          break;
        }
      }
      g = 100000000000000ll;
      for (int j = 0; j < 50; j++) {
        e = d | 1ll << j;
        f = i ^ e;
        if (f > e) swap(f, e);
        if (f > i && f < e) g = min(g, e);
      }
      f = i ^ g;
      if (g <= b) {
        cout << 0 << endl;
        cout << 3 << endl;
        cout << i << " " << f << " " << g << endl;
        return 0;
      }
    }
    if ((a ^ (a + 1)) == 1) {
      cout << 1 << endl;
      cout << 2 << endl;
      cout << a << " " << a + 1 << endl;
    } else {
      cout << 1 << endl;
      cout << 2 << endl;
      cout << a + 1 << " " << a + 2 << endl;
    }
    return 0;
  }
  if ((b - a + 1) >= 10) {
    for (long long i = a; i <= b; i++) {
      if (i % 4ll == 0) {
        cout << 0 << endl;
        cout << 4 << endl;
        cout << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << endl;
        return 0;
      }
    }
  }
  d = 100000000000000ll;
  e = -1;
  for (int i = (0); i < (1 << (b - a + 1)); ++i) {
    if (i == 0) continue;
    if (__builtin_popcount(i) > c) continue;
    f = 0;
    for (int j = (0); j < ((b - a + 1)); ++j) {
      if (i & 1 << j) {
        f ^= (a + (long long)j);
      }
    }
    if (f < d) {
      d = f;
      e = i;
    }
  }
  cout << d << endl;
  cout << __builtin_popcountll(e) << endl;
  for (int i = (0); i < ((b - a + 1)); ++i) {
    if (e & 1 << i) cout << a + (long long)i << " ";
  }
  cout << endl;
}