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() {
  unsigned long long l, r, k, ans;
  cin >> l >> r >> k;
  if (k == 1) {
    cout << l << endl << 1 << endl << l;
  }
  if (k == 2) {
    if (r - l + 1 > 2) {
      cout << 1 << endl << 2 << endl;
      if (l % 2 == 0) {
        cout << l << " " << l + 1 << endl;
        return 0;
      } else {
        cout << l + 1 << " " << l + 2 << endl;
        return 0;
      }
    } else {
      if (l % 2 == 0) {
        cout << 1 << endl << 2 << endl;
        cout << l << " " << l + 1 << endl;
        return 0;
      } else {
        if ((l ^ r) < l)
          cout << (l ^ r) << endl << 2 << endl << l << " " << r;
        else
          cout << l << endl << 1 << endl << l;
      }
    }
  }
  if (k >= 4) {
    if (r - l + 1 > 4) {
      cout << 0 << endl << 4 << endl;
      if (l % 2 == 0) {
        cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3;
        return 0;
      } else if (l % 2 == 1) {
        cout << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4;
        return 0;
      }
    } else {
      if (l % 2 == 0) {
        cout << 0 << endl << 4 << endl;
        cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3;
        return 0;
      } else {
        k = 3;
      }
    }
  }
  if (k == 3) {
    unsigned long long tot = 0;
    unsigned long long ql = l, qr = r;
    for (unsigned long long i = 0;; i++) {
      ql = ql >> 1;
      qr = qr >> 1;
      if (ql == 0) {
        if (qr == 0) ans = 1;
        if (qr != 0) ans = 0;
        tot = i;
        break;
      }
    }
    if (ans == 0) {
      unsigned long long q = 2;
      for (unsigned long long i = 0; i < tot; i++) q = q << 1;
      qr = qr >> 1;
      if (qr > 0) {
        unsigned long long x = q - 1;
        unsigned long long y = q;
        unsigned long long z = (q << 1) - 1;
        cout << 0 << endl << 3 << endl;
        cout << x << " " << q << " " << z;
        return 0;
      }
      if ((q + (q >> 1)) <= r) {
        unsigned long long xx = l;
        unsigned long long yy = q + l - (q >> 1);
        unsigned long long zz = q + (q >> 1);
        cout << 0 << endl << 3 << endl;
        cout << xx << " " << zz << " " << yy;
        return 0;
      } else
        ans = 1;
    }
    if (ans == 1) {
      cout << 1 << endl << 2 << endl;
      if (l % 2 == 0) {
        cout << l << " " << l + 1;
        return 0;
      } else if (l % 2 == 1) {
        cout << l + 1 << " " << l + 2;
        return 0;
      }
    }
  }
}