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 len, sum, ans, C, wri[15], temp[15], i, S, L, R, k, x, z;
inline void DFS(long long now, long long C, long long sum) {
  if (now == R + 1) {
    if (sum >= ans || !C) return;
    len = C;
    ans = sum;
    for (int i = 1; i <= C; i++) wri[i] = temp[i];
    return;
  }
  if (now > R) return;
  DFS(now + 1, C, sum);
  if (C + 1 > k) return;
  temp[C + 1] = now;
  DFS(now + 1, C + 1, sum ^ now);
}
int main() {
  cin >> L >> R >> k;
  if (L == R) {
    cout << L << endl << 1 << endl << R;
    return 0;
  }
  if (R - L <= 4) {
    ans = 999999999999999ll;
    DFS(L, 0, 0);
    cout << ans << endl << len << endl;
    for (i = 1; i <= len; i++) cout << wri[i] << ' ';
    return 0;
  }
  if (k > 3) {
    S = (L & 1) ? L + 1 : L;
    cout << 0 << endl
         << 4 << endl
         << S << ' ' << S + 1 << ' ' << S + 2 << ' ' << S + 3;
    return 0;
  }
  z = 3;
  x = 1;
  while (z <= R && k == 3) {
    if (x >= L) {
      cout << 0 << endl << 3 << endl << x << ' ' << z - 1 << ' ' << z;
      return 0;
    }
    x = x << 1 | 1;
    z <<= 1;
  }
  if (k == 2 || k == 3) {
    S = (L & 1) ? L + 1 : L;
    cout << 1 << endl << 2 << endl << S << ' ' << S + 1;
    return 0;
  }
  if (k == 1 || k == 3) {
    cout << L << endl << 1 << endl << L;
    return 0;
  }
  return 0;
}