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;
int k, cnt;
long long ans, l, r, p[105], q[105];
void dfs(long long x, long long sum, int tot) {
  if (x == r + 1 || tot == k) {
    if (ans > sum) {
      ans = sum;
      cnt = tot;
      for (int i = 1; i <= tot; i++) p[i] = q[i];
    }
    return;
  }
  if (x == r && tot == 0) {
    q[tot + 1] = x;
    dfs(x + 1, sum ^ x, tot + 1);
  } else {
    dfs(x + 1, sum, tot);
    q[tot + 1] = x;
    dfs(x + 1, sum ^ x, tot + 1);
  }
}
int main() {
  scanf("%lld%lld%lld", &l, &r, &k);
  if (r - l + 1 <= 10) {
    ans = 1e15;
    dfs(l, 0, 0);
    printf("%lld\n%d\n", ans, cnt);
    for (int i = 1; i <= cnt; i++) printf("%lld ", p[i]);
    return 0;
  }
  if (k == 1) {
    printf("%lld\n", l);
    puts("1");
    printf("%lld", l);
    return 0;
  }
  if (k == 2) {
    long long x = l;
    if (x % 2 == 1) x++;
    printf("1\n2\n%lld %lld", x, x + 1);
    return 0;
  }
  if (k >= 4) {
    long long x = l;
    if (x % 2 == 1) x++;
    printf("0\n4\n%lld %lld %lld %lld", x, x + 1, x + 2, x + 3);
    return 0;
  }
  for (int i = 40; i >= 1; i--) {
    for (int j = i - 1; j >= 0; j--) {
      long long a = (1LL << i) + (1LL << j);
      long long c = a ^ (a - 1);
      if (l <= c && a <= r) {
        puts("0");
        puts("3");
        printf("%lld %lld %lld", a, a - 1, c);
        return 0;
      }
    }
  }
  long long x = l;
  if (x % 2 == 1) x++;
  printf("1\n2\n%lld %lld", x, x + 1);
  return 0;
}