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 long long INF = 1000000000;
const long long INFLL = INF * INF;
long long pot2[100];
vector<long long> czek_all(long long l, long long r, long long k) {
  vector<long long> wynik;
  long long x = r - l + 1;
  long long res = INFLL;
  for (int i = (1); i <= (pot2[x] - 1); ++i) {
    vector<long long> v;
    long long a = i, ile = 0;
    long long tmp = 0;
    for (int j = (0); j <= ((x)-1); ++j) {
      if (a % 2 == 1) {
        v.push_back(l + j);
        tmp = tmp ^ (l + j);
        ile++;
      }
      a /= 2;
    }
    if (tmp < res && ile <= k) {
      res = tmp;
      wynik = v;
    }
  }
  return wynik;
}
void jebaj() {
  pot2[0] = 1;
  for (int i = (0); i <= ((60) - 1); ++i) pot2[i + 1] = pot2[i] * 2;
  long long l, r, k;
  cin >> l >> r >> k;
  if (k == 1) {
    cout << l << "\n" << 1 << "\n" << l << "\n";
    return;
  }
  if (r - l < 8) {
    vector<long long> v = czek_all(l, r, k);
    long long kk = 0;
    for (__typeof((v).begin()) it = ((v).begin()); it != (v).end(); ++it)
      kk = kk ^ (*it);
    cout << kk << "\n";
    cout << (int)((v).size()) << "\n";
    for (__typeof((v).begin()) it = ((v).begin()); it != (v).end(); ++it)
      cout << *it << " ";
    cout << "\n";
    return;
  }
  long long a = l;
  while (a % 4 > 0) {
    a++;
  }
  if (k >= 4) {
    cout << 0 << "\n" << 4 << "\n";
    for (int i = (0); i <= ((4) - 1); ++i) {
      cout << a << " ";
      a++;
    }
    cout << "\n";
    return;
  }
  if (k == 2) {
    cout << 1 << "\n" << 2 << "\n";
    cout << a << " " << a + 1 << "\n";
    return;
  }
  int h = 0;
  while (pot2[h] <= l) h++;
  long long pot = pot2[h];
  if (pot >= r) {
    cout << 1 << "\n" << 2 << "\n";
    cout << a << " " << a + 1 << "\n";
    return;
  }
  long long x = pot + pot2[h - 1];
  if (x <= r) {
    cout << 0 << "\n" << 3 << "\n";
    cout << pot - 1 << " " << x << " " << (x ^ (pot - 1)) << "\n";
    return;
  }
  cout << 1 << "\n" << 2 << "\n";
  cout << a << " " << a + 1 << "\n";
  return;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  int t;
  t = 1;
  for (int i = (0); i <= ((t)-1); ++i) {
    jebaj();
  }
}