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;
void giveup() {
  L += L % 2;
  cout << "1\n2\n" << L << ' ' << L + 1 << endl;
  exit(0);
}
int main() {
  cin >> L >> R >> K;
  if (K == 1) {
    cout << L << "\n1\n" << L << endl;
    return 0;
  }
  if (K == 2) {
    if (R == L + 1) {
      if (L < (L ^ R)) {
        cout << L << "\n1\n" << L << endl;
      } else {
        cout << (L ^ R) << "\n2\n" << L << ' ' << R << endl;
      }
    } else {
      giveup();
    }
    return 0;
  }
  if (K >= 4) {
    if (L % 2 == 0) {
      cout << "0\n4\n";
      for (int i = 0; i < 4; i++) {
        cout << L + i << ' ';
      }
      return 0;
    }
    if (R - L >= 4) {
      cout << "0\n4\n";
      L++;
      for (int i = 0; i < 4; i++) {
        cout << L + i << ' ';
      }
      return 0;
    }
    int diff = (int)(R - L + 1);
    int maxbit = -1;
    long long minxor = 1ll << 50;
    for (int i = 1, lem = 1 << diff; i < lem; i++) {
      long long curxor = 0;
      for (long long j = 0; j < diff; j++) {
        if (i & (1 << j)) {
          curxor ^= (L + j);
        }
      }
      if (minxor > curxor) {
        minxor = curxor;
        maxbit = i;
      }
    }
    cout << minxor << endl << __builtin_popcount(maxbit) << endl;
    for (long long i = 0; i < diff; i++) {
      if (maxbit & (1 << i)) {
        cout << L + i << ' ';
      }
    }
    return 0;
  }
  for (int i = 0; i < 50; i++) {
    long long lo = (2ll << i) - 1, hi = 3ll << i;
    if (L <= lo && hi <= R) {
      cout << "0\n3\n" << lo << ' ' << hi - 1 << ' ' << hi << endl;
      return 0;
    }
  }
  giveup();
}