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;
vector<long long> getTwo() {
  vector<long long> v;
  long long ret = l;
  v.push_back(l);
  for (long long i = l; i < r; ++i) {
    long long t = i ^ (i + 1);
    if (t < ret) {
      ret = t;
      v.clear();
      v.push_back(i);
      v.push_back(i + 1);
      if (t == 1) {
        break;
      }
    }
  }
  return v;
}
vector<long long> getThree() {
  vector<long long> v;
  for (long long i = 1; i <= 60; ++i) {
    long long a = (1LL << i) | (1LL << (i - 1));
    long long b = a - 1;
    long long c = (1LL << i) - 1;
    if (c >= l && a <= r) {
      v.push_back(c);
      v.push_back(b);
      v.push_back(a);
      break;
    }
  }
  return v;
}
vector<long long> getFour() {
  vector<long long> v;
  for (long long i = l; i + 3 <= r; ++i) {
    long long a = i, b = i + 1, c = i + 2, d = i + 3;
    if ((a ^ b ^ c ^ d) == 0) {
      v.push_back(a);
      v.push_back(b);
      v.push_back(c);
      v.push_back(d);
      break;
    }
  }
  return v;
}
void print(vector<long long> &ans) {
  long long ret = 0;
  for (int i = 0; i < ans.size(); ++i) {
    ret ^= ans[i];
  }
  cout << ret << endl;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int main() {
  cin >> l >> r >> k;
  if (k == 1) {
    vector<long long> ans;
    ans.push_back(l);
    print(ans);
  } else if (k == 2) {
    vector<long long> ans = getTwo();
    print(ans);
  } else if (k == 3) {
    vector<long long> ans = getThree();
    if (ans.size() == 0) {
      ans = getTwo();
    }
    print(ans);
  } else {
    vector<long long> ans = getFour();
    if (ans.size() == 0) {
      ans = getThree();
    }
    if (ans.size() == 0) {
      ans = getTwo();
    }
    print(ans);
  }
  return 0;
}