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;
const int N = 300005;
long long l, r, k;
int main() {
  scanf("%I64d %I64d %I64d", &l, &r, &k);
  if (r - l >= 10) {
    if (k == 1) {
      printf("%I64d\n%d\n%I64d\n", l, 1, l);
      return 0;
    }
    if (k == 2) {
      for (long long s = l; s <= r; s++) {
        long long a = s, b = s + 1;
        if ((a ^ b) == 1) {
          printf("1\n2\n%I64d %I64d\n", a, b);
          return 0;
        }
      }
    }
    if (k == 3) {
      long long key = 0;
      for (int i = 0; i < 60; i++) {
        if ((1LL << i) > l) {
          key = 1LL << i;
          break;
        }
      }
      long long A = l, B = key + (key >> 1), C = A ^ B;
      if (B <= r && C >= l && C <= r) {
        printf("0\n3\n%I64d %I64d %I64d\n", A, B, C);
        return 0;
      }
      for (long long s = l; s <= r; s++) {
        long long a = s, b = s + 1;
        if ((a ^ b) == 1) {
          printf("1\n2\n%I64d %I64d\n", a, b);
          return 0;
        }
      }
      return 0;
    }
    if (k >= 4) {
      for (long long s = l; s <= r; s++) {
        long long a = s, b = s + 1, c = s + 2, d = s + 3;
        if ((a ^ b ^ c ^ d) == 0) {
          printf("0\n4\n%I64d %I64d %I64d %I64d\n", a, b, c, d);
          return 0;
        }
      }
    }
  } else {
    long long best = r + 1;
    int cnt = r - l + 1, ret, pp;
    for (int i = 0; i < (1 << cnt); i++) {
      int tot = 0;
      long long ans = 0;
      for (int j = 0; j < cnt; j++) {
        if (i & (1 << j)) {
          tot++;
          ans ^= (j + l);
        }
      }
      if (tot >= 1 && tot <= k) {
        if (ans < best) {
          best = ans;
          ret = i;
          pp = tot;
        }
      }
    }
    printf("%I64d\n%d\n", best, pp);
    for (int i = 0; i < cnt; i++) {
      if ((1 << i) & ret) {
        printf("%I64d ", l + i);
      }
    }
    return 0;
  }
  return 0;
}