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;
template <typename T>
inline void Read(T &t) {
  int c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar()) f = (c == '-');
  for (t = 0; c >= '0' && c <= '9'; c = getchar())
    t = (t << 3) + (t << 1) + (c ^ 48);
  if (f) t = -t;
}
long long l, r, k;
long long ans, num, chosen[15], a[15];
void dfs(long long now, int step, int limit) {
  if (step > limit) return;
  if (step >= 1) {
    long long res = 0;
    for (int i = 1; i <= step; i++) res ^= a[i];
    if (res < ans) {
      ans = res, num = step;
      for (int i = 1; i <= step; i++) chosen[i] = a[i];
    }
  }
  if (now > r) return;
  dfs(now + 1, step, limit);
  a[step + 1] = now;
  dfs(now + 1, step + 1, limit);
}
int work() {
  ans = (long long)1e16;
  dfs(l, 0, k);
  printf("%I64d\n%I64d\n", ans, num);
  for (int i = 1; i <= num; i++) printf("%I64d ", chosen[i]);
  return 0;
}
int main() {
  Read(l), Read(r), Read(k);
  if (r - l + 1 <= 5) return work();
  if (k == 1)
    printf("%I64d\n1\n%I64d\n", l, l);
  else if (k == 2) {
    long long ans = (l & 1 ? l + 1 : l);
    printf("1\n2\n%I64d %I64d\n", ans, ans + 1);
  } else if (k >= 4) {
    long long ans = (l & 1 ? l + 1 : l);
    printf("0\n4\n%I64d %I64d %I64d %I64d\n", ans, ans + 1, ans + 2, ans + 3);
  } else if (k == 3) {
    for (long long i = 0; i < 40; i++)
      for (long long j = i + 1; j <= 40; j++) {
        long long temp = (1ll << i) + (1ll << j), tmp = temp ^ (temp - 1);
        if (temp <= r && tmp >= l) {
          printf("0\n3\n%I64d %I64d %I64d\n", temp, temp - 1, tmp);
          return 0;
        }
      }
    long long ans = (l & 1 ? l + 1 : l);
    printf("1\n2\n%I64d %I64d\n", ans, ans + 1);
  }
  return 0;
}