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 main() {
  long long l, r, k;
  cin >> l >> r >> k;
  if (r - l < 10) {
    int n = r - l + 1;
    long long ans = 1e13;
    vector<long long> vec;
    for (int i = 1; i < 1 << n; i++) {
      long long c = 0;
      int cnt = 0;
      for (int j = 0; j < n; j++) {
        if (i & 1 << j) {
          c ^= l + j;
          cnt++;
        }
      }
      if (cnt <= k && c < ans) {
        ans = c;
        vec.clear();
        for (int j = 0; j < n; j++)
          if (i & 1 << j) vec.push_back(l + j);
      }
    }
    cout << ans << endl;
    cout << vec.size() << endl;
    for (int i = 0; i < vec.size(); i++) {
      if (i) cout << " ";
      cout << vec[i];
    }
    cout << endl;
  } else if (k < 4) {
    if (k == 3) {
      long long x = 1;
      while (x <= l) x <<= 1;
      x |= x >> 1;
      if (x <= r && (x ^ l) <= r) {
        cout << "0" << endl;
        cout << "3" << endl;
        cout << l << " " << x << " " << (x ^ l) << endl;
        return 0;
      }
    }
    if (k == 1) {
      cout << l << endl;
      cout << "1" << endl;
      cout << l << endl;
      return 0;
    }
    cout << "1" << endl;
    cout << "2" << endl;
    if (l % 2 == 0)
      cout << l << " " << l + 1 << endl;
    else
      cout << l + 1 << " " << l + 2 << endl;
  } else {
    cout << "0" << endl;
    cout << "4" << endl;
    if (l % 2 == 0) {
      cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
    } else {
      cout << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4 << endl;
    }
  }
  return 0;
}