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;
int bit_count(long long x) {
  int res = 0;
  while (x > 0) {
    x = x / 2;
    res++;
  }
  return res;
}
int first_zero(long long x) {
  int last_bit = bit_count(x) - 1;
  for (long long i = last_bit; i >= 0; i--) {
    if ((x & (1ll << i)) == 0) {
      return i;
    }
  }
  return -1;
}
vector<long long> exist(long long l, long long r) {
  long long cnt = bit_count(l);
  long long x = first_zero(l);
  long long a = (1ll << cnt);
  long long b = a + (1ll << (cnt - 1));
  for (int i = cnt - 2; i >= 0; i--) {
    if (l & (1ll << i)) {
      long long p = (1ll << i);
      if (a + p <= r) {
        a += p;
      } else {
        b += p;
      }
    }
  }
  vector<long long> res;
  long long t = a ^ b;
  if (t >= l and t <= r) {
    if (a >= l and a <= r) {
      if (b >= l and b <= r) {
        res.push_back(a);
        res.push_back(b);
        res.push_back(t);
      }
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long l, r, k;
  cin >> l >> r >> k;
  long long res = 1e18;
  vector<long long> ans;
  if (r - l < 10) {
    vector<long long> v;
    for (long long i = l; i <= r; i++) {
      v.push_back(i);
    }
    for (int i = 1; i < (1 << v.size()); i++) {
      vector<long long> st;
      long long cxr = 0;
      for (int j = 0; j < v.size(); j++) {
        if (i & (1 << j)) {
          st.push_back(v[j]);
          cxr = cxr ^ v[j];
        }
      }
      if (cxr < res and st.size() <= k) {
        res = cxr;
        ans = st;
      }
    }
  } else {
    if (k == 1) {
      res = l;
      ans.push_back(l);
    } else if (k == 2) {
      if (l & 1) l++;
      res = 1;
      ans.push_back(l);
      ans.push_back(l + 1);
    } else if (k == 3) {
      if (exist(l, r).size() != 0) {
        res = 0;
        ans = exist(l, r);
      } else {
        if (l & 1) l++;
        res = 1;
        ans.push_back(l);
        ans.push_back(l + 1);
      }
    } else {
      if (l & 1) l++;
      res = 0;
      ans.push_back(l);
      ans.push_back(l + 1);
      ans.push_back(l + 2);
      ans.push_back(l + 3);
    }
  }
  cout << res << '\n';
  cout << ans.size() << '\n';
  for (auto i : ans) {
    cout << i << ' ';
  }
  return 0;
}