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;
const long long int INF = 0x3f3f3f3f3f3f;
long long int l, r, k;
void solve_small(long long int, long long int, long long int);
void solve_big(long long int, long long int, long long int);
int number_ones(int);
int main() {
  cin >> l >> r >> k;
  if (r - l <= 4) {
    solve_small(l, r, k);
  } else {
    solve_big(l, r, k);
  }
  return 0;
}
void solve_small(long long int a, long long int b, long long int sz) {
  long long int max_index = 1 << (b - a + 1);
  long long int ret = INF;
  vector<long long int> ans;
  for (int state = 1; state < max_index; ++state) {
    if (number_ones(state) <= sz) {
      long long int sumXor = 0;
      for (int i = 0; i < b - a + 1; ++i) {
        if ((state >> i) & 1) {
          sumXor ^= a + i;
        }
      }
      if (sumXor < ret) {
        ret = sumXor;
        ans.clear();
        for (int i = 0; i < b - a + 1; ++i) {
          if ((state >> i) & 1) {
            ans.push_back(a + i);
          }
        }
      }
    }
  }
  cout << ret << endl << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int number_ones(int x) {
  int ret = 0;
  while (x) {
    ret += x & 1;
    x >>= 1;
  }
  return ret;
}
void solve_big(long long int a, long long int b, long long int sz) {
  if (sz == 1) {
    cout << a << endl << 1 << endl << a << endl;
    return;
  }
  if (sz == 2) {
    cout << 1 << endl << 2 << endl;
    if (a & 1) {
      cout << a + 1 << " " << a + 2 << endl;
    } else {
      cout << a << " " << a + 1 << endl;
    }
    return;
  }
  if (sz >= 4) {
    cout << 0 << endl << 4 << endl;
    if (a & 1) ++a;
    cout << a << " " << a + 1 << " " << a + 2 << " " << a + 3 << endl;
    return;
  }
  long long int mn = 1;
  long long int mx = 3;
  while (mx <= b) {
    if (mn >= a) {
      cout << 0 << endl << 3 << endl;
      cout << mn << " " << mx - 1 << " " << mx << endl;
      return;
    }
    mn <<= 1;
    ++mn;
    mx <<= 1;
  }
  cout << 1 << endl << 2 << endl;
  if (a & 1) ++a;
  cout << a << " " << a + 1 << endl;
}