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;
void f() {
  vector<long long> o;
  long long va = -1;
  int len = min((long long)(r - l + 1), 8LL);
  for (int i = 1; i < (1 << len); ++i) {
    int tmp = 0;
    for (int j = 0; j < len; ++j) tmp += bool((1 << j) & i);
    if (tmp > k) continue;
    long long s = 0;
    vector<long long> c;
    for (int j = 0; j < len; ++j) {
      if ((1 << j) & i) {
        s ^= l + j;
        c.push_back(l + j);
      }
    }
    if (va == -1 || va > s) va = s, o = c;
  }
  cout << va << endl;
  cout << int(o.size()) << endl;
  for (int j = 0; j < int(o.size()); ++j) {
    cout << o[j] << " ";
  }
}
int getTop(long long x) {
  for (int i = 60; i >= 0; --i) {
    if (x & (1LL << i)) return i;
  }
}
int main() {
  cin >> l >> r >> k;
  long long len = r - l + 1;
  if (k >= 5 || (k == 4 && (l % 2 == 0 || len > 4))) {
    if (l & 1) ++l;
    puts("0\n4");
    int t = 4;
    while (t--) {
      cout << (l++) << " ";
    }
    return 0;
  }
  if (k == 4) {
    f();
    return 0;
  }
  if (k == 3) {
    int top = getTop(l);
    long long x = (1LL << top) ^ (1LL << (top + 1));
    if (x <= r) {
      long long o = l ^ (1LL << top) ^ (1LL << (top + 1));
      puts("0\n3");
      { cout << l << " " << o << " " << x; }
    } else {
      f();
    }
    return 0;
  }
  f();
  return 0;
}