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 INF = 1000000000000000;
int main() {
  long long l, r, k;
  cin >> l >> r >> k;
  if (k == 1)
    cout << l << endl << 1 << endl << l << endl;
  else if (r - l < 4) {
    k = min(k, r - l + 1);
    long long ans = INF, bj = 0;
    int cnta = 0;
    for (int j = 1; j < (1 << (r - l + 1)); j++) {
      int tmp = j;
      long long tans = 0;
      int cnt = 0;
      for (long long i = l; i <= r; i++, tmp >>= 1)
        if (tmp & 1) {
          tans ^= i;
          cnt++;
        }
      if (cnt <= k && tans < ans) {
        cnta = cnt;
        ans = tans;
        bj = j;
      }
    }
    cout << ans << endl;
    cout << cnta << endl;
    for (long long i = l; i <= r; i++, bj >>= 1)
      if (bj & 1) cout << i << " ";
    cout << endl;
  } else {
    if (k == 2) {
      for (l++; l <= r; l++)
        if (l % 2 == 1) {
          cout << 1 << endl << 2 << endl << l << " " << l - 1 << endl;
          break;
        }
    } else if (k == 3) {
      long long t = 1, t2 = l;
      while (t2) {
        t2 /= 2;
        t *= 2;
      }
      if (t + t / 2 <= r)
        cout << 0 << endl
             << 3 << endl
             << l << " " << l + t / 2 << " " << t + t / 2 << endl;
      else
        for (l++; l <= r; l++)
          if (l % 2 == 1) {
            cout << 1 << endl << 2 << endl << l << " " << l - 1 << endl;
            break;
          }
    } else
      for (l++; l <= r; l++)
        if (l % 2 == 1) {
          cout << 0 << endl
               << 4 << endl
               << l + 2 << " " << l + 1 << " " << l << " " << l - 1 << endl;
          break;
        }
  }
}