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, cur;
int k, cc, cnt;
bool ac[4], cho[4];
void exh4() {
  long long ans = 1E13;
  for (int i = (1); i <= (15); i++) {
    cur = cnt = 0;
    for (int j = (0); j <= (3); j++) cho[j] = false;
    for (int j = (0); j <= (3); j++)
      if (i & (1 << j)) cur ^= l + j, cho[j] = true, cnt++;
    if (cur < ans) {
      ans = cur;
      cc = cnt;
      for (int j = (0); j <= (3); j++) ac[j] = cho[j];
    }
  }
  printf("%lld\n%d\n", ans, cc);
  for (int i = (0); i <= (3); i++)
    if (ac[i]) printf("%lld%c", l + i, (--cc) ? ' ' : '\n');
  return;
}
bool poss0() {
  if (r - l < 2) return false;
  long long po = 1, po2;
  while (po * 2 <= r) po *= 2;
  if (l >= po) return false;
  if (r == po) {
    r--;
    return poss0();
  }
  po2 = po / 2;
  while (po2 > 0 && po + po2 > r) po2 /= 2;
  long long res = (po + po2) ^ (po + po2 - 1);
  if (res >= l) {
    printf("0\n3\n%lld %lld %lld\n", po + po2, po + po2 - 1, res);
    return true;
  }
  r = po - 1;
  return poss0();
}
int main() {
  scanf("%lld %lld %d", &l, &r, &k);
  if (k >= 4) {
    if (r - l == 3)
      exh4();
    else {
      if (l & 1) l++;
      printf("0\n4\n%lld %lld %lld %lld\n", l, l + 1, l + 2, l + 3);
    }
    return 0;
  }
  if (k == 2) {
    if (r - l == 1) {
      if (l < (l ^ r))
        printf("%lld\n1\n%lld\n", l, l);
      else
        printf("%lld\n2\n%lld %lld\n", l ^ r, l, r);
    } else {
      if (l & 1) l++;
      printf("1\n2\n%lld %lld\n", l, l + 1);
    }
    return 0;
  }
  if (k == 1) {
    printf("%lld\n1\n%lld\n", l, l);
    return 0;
  }
  if (k == 3) {
    if (poss0())
      return 0;
    else {
      if (l & 1) l++;
      printf("1\n2\n%lld %lld\n", l, l + 1);
    }
    return 0;
  }
}