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;
vector<long long> pt;
int main() {
  long long l, r, k;
  while (cin >> l >> r >> k) {
    long long ans = -1;
    if (k == 1 || l == r) {
      printf("%I64d\n", l);
      printf("%d\n", 1);
      printf("%I64d\n", l);
      continue;
    }
    pt.clear();
    if (r - l == 1) {
      ans = l ^ r;
      if (ans < l) {
        printf("%I64d\n", ans);
        printf("%d\n", 2);
        printf("%I64d %I64d\n", l, r);
      } else {
        printf("%I64d\n", l);
        printf("%d\n", 1);
        printf("%I64d\n", l);
      }
      continue;
    }
    ans = 1;
    if (l & 1) {
      pt.push_back(l + 1);
      pt.push_back(l + 2);
    } else {
      pt.push_back(l);
      pt.push_back(l + 1);
    }
    long long tmp = r;
    int b = 0;
    while (tmp) {
      tmp >>= 1;
      b++;
    }
    tmp = 1ll << (b - 1);
    if (k >= 3)
      for (int i = b - 2; i >= 0; i--) {
        if (r & (1ll << i)) {
          tmp = 0;
          for (int j = i; j >= 0; j--) tmp |= (1ll << j);
          if (tmp >= l) {
            ans = 0;
            pt.clear();
            pt.push_back(tmp);
            pt.push_back(tmp ^ r);
            pt.push_back(r);
          }
          break;
        }
      }
    if (ans && k >= 3) {
      tmp = 1ll << (b - 1);
      tmp--;
      if (tmp / 2 >= l) {
        ans = 0;
        pt.clear();
        pt.push_back(tmp / 2);
        pt.push_back(tmp - (tmp / 2));
        pt.push_back(tmp);
      }
    }
    if (k >= 4 && ans) {
      for (long long i = l; i <= r - 3; i++) {
        long long a = i, b = i + 1, c = i + 2, d = i + 3;
        if ((a ^ b ^ c ^ d) == 0) {
          ans = 0;
          pt.clear();
          pt.push_back(a);
          pt.push_back(b);
          pt.push_back(c);
          pt.push_back(d);
          break;
        }
      }
    }
    cout << ans << endl;
    printf("%d\n", pt.size());
    for (int i = 0; i < pt.size(); i++) {
      printf("%I64d ", pt[i]);
    }
    puts("");
  }
}