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;
int main() {
  ios::sync_with_stdio(0);
  cin >> l >> r >> k;
  if (r - l <= 6) {
    long long x = l;
    int s = r - l + 1;
    for (int i = 0; i < (1 << s); ++i) {
      int p = 0;
      for (int j = 0; j < s; ++j)
        if (i & (1 << j)) p++;
      if (!p || p > k) continue;
      long long c = 0;
      for (int j = 0; j < s; ++j)
        if (i & (1 << j)) c ^= l + j;
      x = min(x, c);
    }
    for (int i = 0; i < (1 << s); ++i) {
      int p = 0;
      for (int j = 0; j < s; ++j)
        if (i & (1 << j)) p++;
      if (!p || p > k) continue;
      long long c = 0;
      for (int j = 0; j < s; ++j)
        if (i & (1 << j)) c ^= l + j;
      if (c == x) {
        cout << x << "\n";
        cout << p << "\n";
        for (int j = 0; j < s; ++j)
          if (i & (1 << j)) cout << l + j << " ";
        cout << "\n";
        return 0;
      }
    }
    return 0;
  }
  if (k == 1) {
    cout << l << "\n";
    cout << 1 << "\n";
    cout << l << "\n";
    return 0;
  }
  if (k == 2) {
    cout << 1 << "\n";
    cout << 2 << "\n";
    if (l % 2) l++;
    cout << l << " " << l + 1 << "\n";
    return 0;
  }
  if (k >= 4) {
    cout << 0 << "\n";
    cout << 4 << "\n";
    if (l % 2) l++;
    cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << "\n";
    return 0;
  }
  long long et = 1;
  for (int i = 0; i < 41; ++i)
    for (int j = 1; j < 50 - i; ++j) {
      int tot = i + j + 1;
      long long x = 0, y = 0, z = 0;
      x = ((1LL << tot) - 1) ^ ((1LL << i) - 1);
      y = ((1LL << tot) - 1) ^ (1LL << i);
      z = ((1LL << (i + 1)) - 1);
      if (x <= r && z >= l) {
        cout << 0 << "\n" << 3 << "\n" << x << " " << y << " " << z << "\n";
        return 0;
      }
    }
  if (l % 2) l++;
  cout << 1 << "\n" << 2 << "\n" << l << " " << l + 1 << "\n";
}