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 <class T>
inline void read(T& x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
int cnt;
long long a[10];
long long l, r, k;
long long bitcnt(long long x) {
  long long ans = 0;
  while (x) ans += x & 1, x >>= 1;
  return ans;
}
long long check(long long x) {
  long long ans = 0;
  for (int i = 1; i <= cnt; i++)
    if (x & (1 << (i - 1))) ans ^= a[i];
  return ans;
}
void force() {
  cnt = 0;
  long long ans = (1LL << 60), pos;
  for (long long i = l; i <= r; i++) a[++cnt] = i;
  for (int i = 1; i < (1 << cnt); i++) {
    if (bitcnt(i) <= k && check(i) < ans) {
      ans = check(i);
      pos = i;
    }
  }
  printf("%I64d\n", ans);
  printf("%I64d\n", bitcnt(pos));
  for (int i = 1; i <= cnt; i++)
    if (pos & (1 << (i - 1))) printf("%I64d ", a[i]);
}
int main() {
  read(l), read(r), read(k);
  if (r - l <= 6)
    force();
  else {
    if (k == 1) {
      printf("%I64d\n", l);
      printf("1\n");
      printf("%I64d\n", l);
    }
    if (k == 2) {
      long long p = l / 2 + 1;
      printf("1\n");
      printf("2\n");
      printf("%I64d %I64d\n", p * 2, p * 2 + 1);
    }
    if (k == 3) {
      long long x = 6, y = 5, z = 3;
      while (z <= r) {
        if (l <= z && x <= r) {
          printf("0\n");
          printf("3\n");
          printf("%I64d %I64d %I64d\n", x, y, z);
          return 0;
        }
        x = x << 1;
        y = (y << 1) + 1;
        z = (z << 1) + 1;
      }
      long long p = l / 2 + 1;
      printf("1\n");
      printf("2\n");
      printf("%I64d %I64d\n", p * 2, p * 2 + 1);
    }
    if (k >= 4) {
      long long p = l / 2 + 1;
      printf("0\n");
      printf("4\n");
      printf("%I64d %I64d ", p * 2, p * 2 + 1);
      p++;
      printf("%I64d %I64d\n", p * 2, p * 2 + 1);
    }
  }
  return 0;
}