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 int N = 100005;
const int M = 5005;
const int INF = 1000000007;
long long n, m, w, mini, k, l, r, anss;
vector<long long> ans, t;
void opt() {
  long long s = 0;
  for (int i = 0; i < t.size(); i++) s ^= t[i];
  if ((ans.size() == 0 || s < anss) && t.size() > 0) {
    ans = vector<long long>(t);
    anss = s;
  }
}
void tryplus(long long x) {
  if (x > r || t.size() == k) {
    opt();
    return;
  }
  tryplus(x + 1);
  t.push_back(x);
  tryplus(x + 1);
  t.erase(t.end() - 1);
}
int main() {
  cin >> l >> r >> k;
  long long j = 0;
  long long d = 1;
  for (long long i = 0; (d << i) <= r; i++)
    if ((d << i) & r) j = i;
  if (r - (d << j) >= l)
    if (k >= 3) {
      t = vector<long long>();
      t.push_back(r - (d << j));
      t.push_back((d << j));
      t.push_back(r);
      opt();
    }
  for (long long i = 0; (d << i) <= l; i++)
    if ((d << i) & l) j = i;
  j++;
  if (l + (d << j) <= r)
    if (k >= 3) {
      t = vector<long long>();
      t.push_back(l);
      t.push_back((d << j));
      t.push_back(l + (d << j));
      opt();
    }
  d = l ^ r;
  if (d >= l && d <= r)
    if (k >= 3) {
      t = vector<long long>();
      t.push_back(l);
      t.push_back(d);
      t.push_back(r);
      opt();
    }
  if (r - l > 10) r = l + 10;
  tryplus(l);
  long long s = 0;
  for (int i = 0; i < ans.size(); i++) s ^= ans[i];
  cout << s << endl << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}