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 = 200000;
long long l, r, m;
void into() { scanf("%lld%lld%lld", &l, &r, &m); }
vector<long long> ans;
void work() {
  if (r - l + 1 <= 6) {
    long long mn = l;
    ans = {l};
    for (int s = 1; s < 1 << r - l + 1; ++s) {
      long long now = 0;
      int cnt = 0;
      for (int i = 0; i <= r - l; ++i)
        cnt += s >> i & 1, now ^= (s >> i & 1) * (l + i);
      if (cnt && cnt <= m && now < mn) {
        mn = now;
        ans.clear();
        for (int i = 0; i <= r - l; ++i)
          if (s >> i & 1) ans.push_back(l + i);
      }
    }
    return;
  }
  if (m == 1) {
    ans = {l};
    return;
  }
  if (m == 2) {
    ans = {l + (l & 1), l + (l & 1) + 1};
    return;
  }
  if (m >= 4) {
    ans = {l + (l & 1), l + (l & 1) + 1, l + (l & 1) + 2, l + (l & 1) + 3};
    return;
  }
  long long t = 0;
  for (int i = 60; i >= 0; --i)
    if (l >> i & 1) {
      t = i;
      break;
    }
  t = 1LL << t;
  ans = t * 3 <= r ? (vector<long long>){l, l + t, t * 3}
                   : (vector<long long>){l + (l & 1), l + (l & 1) + 1};
}
void outo() {
  long long xr = 0;
  for (long long x : ans) xr ^= x;
  printf("%lld\n%d\n", xr, ans.size());
  for (long long x : ans) printf("%lld ", x);
  puts("");
}
int main() {
  int T = 1;
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}