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>
long long l, r, k, a, b, c;
bool gao(long long x) {
  long long y = (x >> 1);
  while (y) {
    c = x + y;
    b = c - 1;
    a = b ^ c;
    if (l <= a && c <= r) return true;
    y >>= 1;
  }
  return false;
}
using namespace std;
int main() {
  scanf("%I64d%I64d%I64d", &l, &r, &k);
  if (r - l < 4) {
    int cnt = r - l + 1;
    int best = 0;
    long long MIN = (long long)1e12;
    for (int i = 1; i < (1 << cnt); i++) {
      if (__builtin_popcount(i) > k) continue;
      long long t = 0LL;
      for (int j = 0; j < cnt; j++) {
        if (i >> j & 1) {
          t ^= (l + j);
        }
      }
      if (t < MIN) {
        MIN = t;
        best = i;
      }
    }
    printf("%I64d\n%d\n", MIN, __builtin_popcount(best));
    for (int j = 0; j < cnt; j++) {
      if (best >> j & 1) printf("%I64d ", l + j);
    }
    printf("\n");
  } else {
    if (k == 1) {
      printf("%I64d\n1\n%I64d\n", l, l);
    } else if (k == 2) {
      if (l & 1) l++;
      printf("1\n2\n%I64d %I64d\n", l, l + 1);
    } else if (k == 3) {
      long long t = 1LL;
      bool flag = 0;
      for (; t < r && !flag; t <<= 1) {
        if (t > l) flag = gao(t);
      }
      if (flag) {
        printf("0\n3\n%I64d %I64d %I64d\n", a, b, c);
      } else {
        if (l & 1) l++;
        printf("1\n2\n%I64d %I64d\n", l, l + 1);
      }
    } else if (k >= 4) {
      if (l & 1) l++;
      printf("0\n4\n");
      for (long long i = l; i < l + 4; i++)
        printf("%I64d%c", i, (i == l + 3) ? '\n' : ' ');
    }
  }
  return 0;
}