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;
long long l, r, k;
vector<long long> ans;
void print() {
  long long x = 0;
  for (int i = 0; i < ans.size(); i++) x ^= ans[i];
  cout << x << endl;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  exit(0);
}
void brute() {
  int x = r - l + 1;
  long long min = LLONG_MAX;
  for (int mask = 1; mask < 1 << x; mask++) {
    int net = __builtin_popcount(mask);
    if (net > k) continue;
    long long here = 0;
    for (int bit = 0; bit < x; bit++) {
      if (mask & (1LL << bit)) here = here ^ (l + bit);
    }
    if (here < min) {
      min = here;
      ans.clear();
      for (int bit = 0; bit < x; bit++)
        if (mask & (1LL << bit)) ans.push_back(l + bit);
    }
  }
  print();
}
void yolo1() {
  ans.push_back(l);
  print();
}
void yolo2() {
  if (l % 2) l++;
  ans.push_back(l);
  ans.push_back(l + 1);
  print();
}
void yolo3() {
  long long bit = 0, ll = l;
  while (l > 0) l /= 2, bit++;
  long long no = (1LL << bit);
  no += no / 2, l = ll;
  if (no <= r) {
    ans.push_back(no);
    ans.push_back(l);
    ans.push_back(l ^ no);
    print();
  }
  yolo2();
}
void yolo4() {
  if (l % 2) l++;
  ans.push_back(l);
  ans.push_back(l + 1);
  ans.push_back(l + 2);
  ans.push_back(l + 3);
  print();
}
int main() {
  cin >> l >> r >> k;
  if (r - l <= 6) brute();
  if (k == 1) yolo1();
  if (k == 2) yolo2();
  if (k == 3) yolo3();
  yolo4();
  return 0;
}