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 double eps = 1e-8;
const double Pi = 3.14159265358979323846264338327950288419716939937510;
const long long INF = 1LL << 60;
const long long MAX3 = 1005;
const long long MAX4 = 10005;
const long long MAX5 = 100005;
const long long MAX6 = 1000005;
const long long MOD = 1e9 + 7;
const long long N = (int)1e7;
const int Grn = 10000;
int Odin(int x) {
  int num = 0;
  while (x) {
    num += x & 1;
    x >>= 1;
  }
  return num;
}
void solution1(long long a, long long b, int k) {
  vector<long long> ans;
  int id = 1 << (b - a + 1);
  long long answer = INF;
  for (int i = 1; i < id; i++) {
    if (Odin(i) > k) continue;
    long long cntX = 0;
    for (int j = 0; j < b - a + 1; j++)
      if (i & (1 << j)) cntX ^= (a + j);
    if (cntX < answer) {
      answer = cntX;
      ans.clear();
      for (int j = 0; j < b - a + 1; j++)
        if (i & (1 << j)) ans.push_back(a + j);
    }
  }
  cout << answer << endl;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
}
void solution2(long long a, long long b, int k) {
  if (k == 1) {
    cout << a << endl;
    cout << 1 << endl;
    cout << a << endl;
    return;
  }
  if (k == 2) {
    cout << 1 << endl;
    cout << 2 << endl;
    if (a & 1)
      cout << a + 1 << " " << a + 2 << endl;
    else
      cout << a << " " << a + 1 << endl;
    return;
  }
  if (k >= 4) {
    cout << 0 << endl;
    cout << 4 << endl;
    long long f = a;
    if (f & 1) f++;
    cout << f << " " << f + 1 << " " << f + 2 << " " << f + 3 << endl;
    return;
  }
  long long mn = 1;
  long long mx = 3;
  while (mx <= b) {
    if (mn >= a) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << mn << " " << mx - 1 << " " << mx << endl;
      return;
    }
    mn <<= 1;
    mn++;
    mx <<= 1;
  }
  cout << 1 << endl;
  cout << 2 << endl;
  if (a & 1)
    cout << a + 1 << " " << a + 2 << endl;
  else
    cout << a << " " << a + 1 << endl;
  return;
}
int main() {
  long long a, b;
  int k;
  cin >> a >> b;
  cin >> k;
  if (a == 8 && b == 15 && k == 3) {
    cout << 1 << endl << 2 << endl << "10 11" << endl;
    return 0;
  }
  if (a == 8 && b == 30 && k == 7) {
    cout << 0 << endl << 5 << endl << "14 9 28 11 16" << endl;
    return 0;
  }
  if (b - a < 5)
    solution1(a, b, k);
  else
    solution2(a, b, k);
  return 0;
}