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 maxn = 100005;
const long long inf = 1ll << 60;
const int mod = 1000000007;
const double eps = 1e-8;
long long l, r;
int k;
int bit(long long x) {
  int res = 0;
  long long tp = 1;
  while (tp * 2 <= x) {
    res++;
    tp *= 2;
  }
  return res;
}
void sol() {
  int bl = bit(l), br = bit(r);
  long long tp = 1ll << bl;
  if (bl == br || 3 * tp > r) {
    puts("1");
    puts("2");
    if (l % 2 == 0) {
      printf("%lld %lld\n", l, l + 1);
    } else {
      printf("%lld %lld\n", l + 1, l + 2);
    }
  } else {
    puts("0");
    puts("3");
    printf("%lld %lld %lld\n", 2 * tp - 1, 3 * tp - 1, 3 * tp);
  }
}
int main() {
  scanf("%lld%lld%d", &l, &r, &k);
  long long all = r - l + 1;
  if (all >= 7) {
    if (k >= 4) {
      for (long long i = l;; i++) {
        if (i % 4 == 0) {
          puts("0");
          puts("4");
          printf("%lld %lld %lld %lld\n", i, i + 1, i + 2, i + 3);
          return 0;
        }
      }
    } else if (k == 1) {
      printf("%lld\n1\n%lld\n", l, l);
    } else if (k == 2) {
      for (long long i = l;; i++) {
        if (i % 2 == 0) {
          printf("1\n2\n%lld %lld\n", i, i + 1);
          return 0;
        }
      }
    } else {
      sol();
    }
  } else {
    long long res = inf;
    vector<long long> v;
    for (int i = 1; i < (1 << all); i++) {
      int cn = 0;
      long long tp = 0;
      for (int j = 0; j < all; j++)
        if (i & (1 << j)) {
          cn++;
          tp ^= (l + j);
        }
      if (cn <= k && tp < res) {
        res = tp;
        v.clear();
        for (int j = 0; j < all; j++)
          if (i & (1 << j)) {
            v.push_back(l + j);
          }
      }
    }
    printf("%lld\n%lu\n", res, v.size());
    for (int i = 0; i < v.size(); i++) printf("%lld ", v[i]);
  }
}