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;
mt19937 rng(423978);
const long long mod = 1e9 + 7;
const int mn = 3e5 + 10;
vector<long long> v;
int main() {
  long long l, r, k;
  scanf("%lld%lld%lld", &l, &r, &k);
  if (k == 1) {
    printf("%lld\n1\n%lld", l, l);
    return 0;
  }
  long long ans = 0x3f3f3f3f3f3f3f3f;
  if (r - l >= 2 || l % 2 == 0) {
    ans = 1;
  }
  if (k >= 3) {
    long long np = 1LL << 64 - __builtin_clzll(l);
    long long nx = np | (np >> 1);
    if (nx <= r) {
      printf("0\n3\n%lld %lld %lld", l, np + l - (np >> 1), nx);
      return 0;
    }
  }
  if (k >= 4 && (r - l >= 4 || (r - l >= 3 && l % 2 == 0))) {
    if (l % 2 == 1) l++;
    printf("0\n4\n%lld %lld %lld %lld", l, l + 1, l + 2, l + 3);
    return 0;
  }
  if (ans == 1) {
    if (l % 2 == 1) l++;
    printf("1\n2\n%lld %lld", l, l + 1);
    return 0;
  }
  if (l == r || l <= (l ^ r))
    printf("%lld\n1\n%lld", l, l);
  else
    printf("%lld\n2\n%lld %lld", l ^ r, l, r);
}