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 n, k, l, r, calc, cnt, ans = 2e15, mask;
int main() {
  cin >> l >> r >> k;
  n = r - l + 1;
  if (n <= 20) {
    for (long long i = 1; i < (1 << n); i++) {
      calc = cnt = 0;
      for (long long j = 0; j < n; j++)
        if ((i >> j) & 1) cnt++, calc ^= l + j;
      if (calc < ans && cnt <= k) ans = calc, mask = i;
    }
    cout << ans << endl;
    cnt = 0;
    for (long long i = 0; i < n; i++) cnt += ((mask >> i) & 1);
    cout << cnt << endl;
    for (long long i = 0; i < n; i++)
      if ((mask >> i) & 1) cout << l + i << " ";
    return 0;
  }
  if (k == 1) {
    cout << l << "\n1\n" << l;
    return 0;
  }
  if (k == 2) {
    cout << "1\n2\n";
    if (l % 2 == 0)
      cout << l << " " << l + 1;
    else
      cout << l + 1 << " " << l + 2;
    return 0;
  }
  if (k == 3) {
    cnt = 2;
    while (cnt + cnt / 2 <= r) {
      if (cnt - 1 >= l) {
        cout << "0\n3\n"
             << cnt + cnt / 2 << " " << cnt + cnt / 2 - 1 << " " << cnt - 1;
        return 0;
      }
      cnt *= 2;
    }
    cout << "1\n2\n";
    if (l % 2 == 0)
      cout << l << " " << l + 1;
    else
      cout << l + 1 << " " << l + 2;
    return 0;
  }
  if (k >= 4) {
    cout << "0\n4\n";
    if (l % 2 == 1)
      cout << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4;
    else
      cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3;
    return 0;
  }
}