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;
template <typename T>
void readV(vector<T> &v, int n) {
  for (int i = 0; i < n; i++) {
    T x;
    cin >> x;
    v.push_back(x);
  }
}
long long next(long long x) {
  int s = 0;
  while (x) {
    s++;
    x >>= 1;
  }
  return s - 1;
}
long long mx(long long x) {
  int n = next(x);
  return (1LL << (n + 1)) + (1LL << n);
}
int main(void) {
  ios::sync_with_stdio(false);
  long long l, r, sz;
  int k;
  scanf("%lld%lld%d", &l, &r, &k);
  sz = r - l + 1;
  vector<long long> ans;
  if (k == 1)
    ans.push_back(l);
  else if (k == 2) {
    if (sz > 2 && ((l ^ (l + 1)) != 1)) l++;
    ans.push_back(l);
    if (l > (l ^ (l + 1))) ans.push_back(l + 1);
  } else if (k == 3) {
    if (mx(l) <= r) {
      ans.push_back(l);
      ans.push_back(mx(l));
      ans.push_back(l ^ mx(l));
    } else {
      if ((l ^ (l + 1)) != 1) l++;
      ans.push_back(l);
      ans.push_back(l + 1);
    }
  } else if (k == 4) {
    if (sz > 4 && (l ^ (l + 1)) != 1) l++;
    if ((l ^ (l + 1)) == 1) {
      ans.push_back(l);
      ans.push_back(l + 1);
      ans.push_back(l + 2);
      ans.push_back(l + 3);
    } else {
      if (mx(l) <= r) {
        ans.push_back(l);
        ans.push_back(mx(l));
        ans.push_back(l ^ mx(l));
      } else {
        if ((l ^ (l + 1)) != 1) l++;
        ans.push_back(l);
        ans.push_back(l + 1);
      }
    }
  } else {
    if ((l ^ (l + 1)) != 1) l++;
    ans.push_back(l);
    ans.push_back(l + 1);
    ans.push_back(l + 2);
    ans.push_back(l + 3);
  }
  long long xo = 0;
  for (int i = 0; i < ans.size(); i++) xo ^= ans[i];
  printf("%lld\n", xo);
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%lld ", ans[i]);
  printf("\n");
}