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;
int main() {
  long long l, r, k;
  cin >> l >> r >> k;
  long long dis = r - l + 1;
  if (k == 1 || dis == 1) {
    cout << l << endl;
    cout << 1 << endl;
    cout << l << endl;
    return 0;
  }
  if (dis == 2) {
    long long num = l ^ r;
    if (num < l) {
      cout << num << endl;
      cout << 2 << endl;
      cout << l << ' ' << r << endl;
    } else {
      cout << l << endl;
      cout << 1 << endl;
      cout << l << endl;
    }
    return 0;
  }
  if (dis > 2 && k == 2) {
    cout << 1 << endl;
    cout << 2 << endl;
    long long x = l;
    if (x & 1) x++;
    cout << x << " " << x + 1 << endl;
    return 0;
  }
  if (k >= 3) {
    long long temp = 1;
    long long x1 = -1;
    for (int i = 0; i < 60; i++) {
      x1 = (temp << i);
      if (x1 > l) break;
    }
    long long x2 = x1 + (x1 >> 1);
    long long x3 = l ^ x2;
    if (x2 >= l && x2 <= r && x3 >= l && x3 <= r) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << l << " " << x2 << ' ' << x3 << endl;
      return 0;
    }
  }
  if (k == 3 && dis >= 3) {
    if (l == 1) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << "1 2 3" << endl;
    } else {
      long long x = l;
      if (x & 1) x++;
      cout << 1 << endl;
      cout << 2 << endl;
      cout << x << " " << x + 1 << endl;
    }
    return 0;
  }
  if (dis == 3 && k > 2) {
    if (l == 1) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << "1 2 3" << endl;
    } else {
      long long x = l;
      if (x & 1) x++;
      cout << 1 << endl;
      cout << 2 << endl;
      cout << x << " " << x + 1 << endl;
    }
    return 0;
  }
  if (dis == 4 && k >= 4) {
    if (l == 1) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << "1 2 3" << endl;
    } else if (l % 2 == 0) {
      cout << 0 << endl;
      cout << 4 << endl;
      cout << l << ' ' << l + 1 << ' ' << l + 2 << ' ' << l + 3 << endl;
    } else {
      long long x = l;
      if (x & 1) x++;
      cout << 1 << endl;
      cout << 2 << endl;
      cout << x << " " << x + 1 << endl;
    }
    return 0;
  }
  cout << 0 << endl;
  cout << 4 << endl;
  long long x = l;
  if (x & 1) x++;
  cout << x << " " << x + 1 << ' ' << x + 2 << ' ' << x + 3 << endl;
  return 0;
}