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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
long long l, r;
set<long long> ans;
bool cal3() {
  int i = 60;
  for (; i >= 0; i--)
    if ((1LL << i) & l) {
      break;
    }
  long long top = (1LL << (i + 1)) + (1LL << i);
  if (top > r) return 0;
  ans.insert(l);
  ans.insert(top);
  ans.insert((1LL << (i + 1)) + l - (1LL << i));
  return 1;
}
void cal() {
  if (k >= 3) {
    if (cal3()) return;
  }
  vector<long long> v;
  for (long long i = l; i <= min(l + 10, r); i++) v.push_back(i);
  n = v.size();
  long long tAns = (1LL << 60);
  for (int s = 1; s < (1 << n); s++) {
    if (__builtin_popcount(s) <= k) {
      long long tp = 0;
      set<long long> gg;
      for (int j = 0; j < n; j++)
        if ((1 << j) & s) {
          tp ^= v[j];
          gg.insert(v[j]);
        }
      if (tAns > tp) {
        tAns = tp;
        ans = gg;
      }
    }
  }
}
int main() {
  cin >> l >> r >> k;
  cal();
  long long g = 0;
  for (long long x : ans) g ^= x;
  printf("%lld\n%d\n", g, ans.size());
  for (long long x : ans) printf("%lld ", x);
  return 0;
}