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;
long long l, r, ans[10];
int k, an = 0;
void build() {
  if (k == 1)
    ans[an++] = l;
  else if (k == 2) {
    if (r - l + 1 == 2) {
      if (l < (l ^ r))
        ans[an++] = l;
      else {
        ans[an++] = l;
        ans[an++] = r;
      }
    } else {
      if (l % 2) l++;
      ans[an++] = l;
      ans[an++] = l + 1;
    }
  } else if (k == 3) {
    long long p = 1;
    while (p <= l) p *= 2;
    p /= 2;
    if (3 * p <= r) {
      ans[an++] = l;
      ans[an++] = l + p;
      ans[an++] = 3 * p;
    } else {
      if (l % 2) l++;
      ans[an++] = l;
      ans[an++] = l + 1;
    }
  } else if (k == 4) {
    if (r - l + 1 > 4) {
      if (l % 2) l++;
      for (an = 0; an < 4; an++) ans[an] = l + an;
    } else {
      long long Min = l;
      ans[an++] = l;
      for (int i = 1; i < 16; i++) {
        long long sum = 0;
        for (int j = 0; j < 4; j++) {
          if (i & (1 << j)) sum ^= (l + j);
        }
        if (sum < Min) {
          Min = sum;
          an = 0;
          for (int j = 0; j < 4; j++) {
            if (i & (1 << j)) ans[an++] = (l + j);
          }
        }
      }
    }
  } else {
    if (l % 2) l++;
    for (an = 0; an < 4; an++) ans[an] = l + an;
  }
}
int main() {
  scanf("%lld%lld%d", &l, &r, &k);
  build();
  long long sum = 0;
  for (int i = 0; i < an; i++) sum ^= ans[i];
  printf("%lld\n", sum);
  printf("%d\n", an);
  for (int i = 0; i < an - 1; i++) printf("%lld ", ans[i]);
  printf("%lld\n", ans[an - 1]);
  return 0;
}